Scribe for April 26

Hello 9-05 welcome back from a long break. Well now it's time for more school and yes we had math today.

Today we talked about question 21 in section 8.3 (textbook). The question went a little something like this......

A square picture is made out of wood that is 1.6 cm wide. The perimeter of the outside of the frame is 75.2 cm. What is the side length of the largest square picture that the frame will display?

Well really what the question is asking you is what the side length of the middle square is.

Now to solve this you have to look at the width first.

Since you know that the width is 1.6 cm and it will also be 1.6 cm on the other side between the small square and the perimeter, you can multiply 1.6 cm by two.

Once you get your answer which is 3.2, you have to divide 75.2 (perimeter) by 4 so you could get the side length each side of the outside of the frame. You want to do this because if you multiply 3.2 (width times 2) by the side length of the outside which 18.2 you will get the side length of the inside square.

Did you get an answer of 15.6? Well if you did then you have got the right answer.

After solving that question we went onto question 27 also in section 8.3 (textbook) which was even difficult for Mr. Backe to solve. Well here is the question...

Tahir is training for an upcoming cross-country meet. He runs 13 km, three times a week. His goal is to increase his average speed by 1.5 km/h, so that he can complete each run in 1 1/4. How long does he take to complete each run now, to the nearest tenth of a minute. Solve this problem in two different ways.

Well I am not really sure how to solve this but this is what Mr.Backe did.

Thats all for today and sorry it is so late. Bye.

Jounel Entry Post

March, 3. 2010

Today in math class we reviewed some homework from yesterday. Next we started on a new unit on multiplying and dividing polynomials.

The first thing that we talked about is how to use algebra tiles to find out the answer to an equation. eg. (2x)(3x). To do this question using algebra tiles, you will first need to know that.....

Now that you know this, lay out your tiles as so.

Your first number in a multiplication equation will always be on the the left part of the diagram. Now just fill in the middle part of your diagram with as many x squared tiles as needed. In this case you have to add 6 x squared tiles.

The tiles in the middle of your diagram is your answer which is 6x squared.

After that we learned how to solve the same equation but algebraically this time. To solve this algebraically you just have to simply multiply both coefficients as well as the variables together to get your answer.

Next we learned how to solve division questions with algebra tiles. For this example we will use this simple question. 6x^2/3x. To solve this question lay out our tiles like this...

Notice that 3x is placed on the top of the diagram instead of the left.

The next step is very simple. Just draw as many tiles as you need so that you can multiply 3x by a certain number to get the the middle amount of tiles which is 6x^2. For this question the missing number is 2x.

To solve this question algebraically you have to divide the coefficients, and variables together. Remember that any variable without a degree will always have a degree of one. Also remember that when dividing the variables together you can subtract the exponents from both of the variables to get your answer.

March 11 Scribe

Hello 9-05 today was pretty much a work period for us to catch up on some of the assignments due next week. So if you didn't know by now I should advise you that there is a TEST NEXT WEDNESDAY on chapter 5 and 7.

Here are other things that are due:

Tuesday- All extra practise and all workbook for both chapter 5 and 7.

Friday- Stash it.

In your stash it there should be:

• Chapter 5 stuff that was originally due before
• Chapter 7 text book mathlinks and wrap it up
• 2 or 3 tests signed

Remember to get the stash-it done because this is worth 20% of your final math mark for chapter 5 and 7.



Okay I think that I am about done so if I made any mistakes just comment.

Scribe for January 25

Hello people that I know from my school, today Mr Backe wasn't here so we had math class in our homeroom. The whole class was pretty much a work period.

So YAY! Isn't this convenient for you guys stuck at home because of the snow or sickness.

If you DON'T REMEMBER THE HOMEWORK from last time we had to do....

• 4.2 The whole section in our mathlinks hard coverbooks.


• The worksheets from before if you didn't finish (the ones about volume ratios).


• The foldy thing work sheet the one where there is work that you have to do and you have to fold the paper.


• Any other textbook work that you didn't finish from before.


• The chapter 4 introduction sheet.

Well thats all for now and if I have forgoten something about the homework from last time just comment because I am a sucky mc baby for not remembering.

The next scribe will be ANYBODY THAT HASN'T DONE A SECOND SCRIBE YET because I am being nice and not picking on anybody!

Dont even think about it nicky!

Merry Christmas and We Miss You

Hey Mr. Backe Jai Abang. Just want you to know how much we all miss you and also merry crhistmas! Will you be coming back soon because i think a good math teacher is only allowed to teach a class like us.



HEY LOOK ITS MEE!!! ->



















Watch this video.





Are you done? Ok then here we go!

What do you call a group of students that really misses Mr. Backe?

THE CLASS (=

What happens if you are away for too long?

Our school days become boring =(

What do you get when you mix school with Mr. Backe?

A school of happy students (=

Now that's over time for some tic tac toe. I will be X and you will be O.

Ok now lets play!!!

The winner is Mr.Backe!

WOW you really are a good teacher (=

Phew... I am done for the day. I guess you have shown yourself worthy.

COME BACK SOON......WE MISS YOU

How to Solve Question # 22

Hi 9-05 it's Jai here coming to you with question #22. For those of you who didn't understand this question I will try to explain it to you so next time that you see question #22 you can say "Hey I know how to solve this!"

The question goes something like this......

Taj has three scoops for measuring flour. The largest scoop holds 2 1/2 as much as the smallest one. The middle scoop holds 1 3/4 as much as the smallest one.

Describe 2 different ways in which Taj could measure each of the following quantities. He can only use full scoops.

a) 3 1/4 as much as the smallest one.
b) 1/2 as much as the smallest one.

So to figure out question (a) out you are going to have to have the following things. Empty cups of each measuring cup, a flour bag, and any container that is big enough to hold more than 3 1/4.

Once you have completed this you have to now fill up the largest measuring cup.

After that fill up the middle measuring cup by pouring as much flour as you can from the largest cup into the middle cup. Once you have done this the largest cup will have 3/4 of flour left.

Once this is done, pour the flour from the largest measuring cup that is left into the random container. Once this is completed. Empty all of the measuring cups into the flour bag and repeat this process 2 more times ensuring that the random container is filled with 2 1/4 as much flour as the smallest measuring cup.

Now the last step into solving question (a) is to fill up the smallest measuring cup and pouring that flour into the random container filling it up to 3 1/4 which question (a) asks for.

Now that we have solved question (a) on to the easiest question to solve which is question (b). To solve this question all you have to do is fill up the largest measuring cup first. Then you empty that measuring cup twice into the smallest measuring cup. Each time that you fill up the smallest measuring cup remember to empty it into the flour bag afterwards.

Once this is completed, the amount of flour that you will have left in the largest measuring cup will be the amount that question (b) asks which is 1/2 as much as the smallest measuring cup.

If you still don't understand what I am trying to say, just ask me at school, and I will try to explain this question even better than I have in this post. Comment if I made any mistakes any mistakes. Thanks for reading!

Question #22

Hi everyone, today I am going to help you solve question number 22 in our math links book for section 2.2. The question goes something like this....

An aircraft is flying at an altitude of 2950 m. It decsended for 3 minutes 2.5 m/s and then decsended for 2.5 minutes at 2.8 m/s. What was the plains altitude after the decsent.

So to start off, we first have to look at how long it decsended, and how many metres it went down each second it went down. For this question we know that it decended two times. For it's first decsend it went down for 3 minutes and for each second in that minute it went down 2.5 meteres.

Now that we know how long it decsended for in its first decsend, and how many metres it went down for each second in that minute we have to find out how many seconds are in 3 minutes. Once we have determined the amount of seconds 3 minutes consists of, we are going to multiply that answer by how many metres in went down per second. In this case we have to multiply 180 by 2.5 to give us an answer of 450.

Once we have finished with the first decsent, we have to remember that we have another decsent still waiting for our answer. For this descent we have to find the amount of seconds in 2 and a half minutes. To find that out you add 120 (2 minutes) by 30 (half of a minute) to get an answer of 150 seconds. Now you just simply multiply 150 by the amount of metres each second the plain decsended by. Your equastion should look somthing like this. 150 x 2.8.

Once you have solved that equastion, just easily add both of your answers together to get your final answer. Your final answer should be 2080 m.

That's all I have to say for now and hope that you understood what I was telling you. Please comment if I have made any mistakes.... Thanks for reading!

Scribe Post for October 13

Hey everybody. Welcome to the scribe post for October 13, 2009.

Today in math class, we reviewed some of the questions that we didn't understand from our homework on Monday. One of the questions we that we checked over was CYU question 1b. In this question it tells us to make 2 different solid objects using 24 interlocking cubes. You then have to slide the two objects together, and identify the area of overlap between the two objects.

To find the area between the overlap of two objects, take the area of the object that has the smaller height and width and multiply that by 2. In this case you have to multiply 4 by 2 to get your answer of 8.

After we were done checking over our h.w from last class, we were given a shape that we had to find the exposed surface area on. Remember that exposed means only the faces that you would be able to see if you were to walk around the whole object. This is very important because it will be on our test tommorow!

Here is the shape and the measurements.

To find out the surface area for this shape, you are first going to need to find all of the measurements you possibally can with these given side lengths.

The first thing that I did was find out the measurement from the dotted line in the triangle, to the edge of the triangle.

To find out this measurement of this side lentgh I divided the base mesurement by 2 to get me an answer of 6.

After I got an answer of 6, I needed to find out the side length of the dotted line from the base of the triangle to the top of the shape.

To do that I subtracted The height by the base to get me an answer of six again. I did this because the dotted line is equal to 15 and 9 cm is already being used by the height. Since there is only 6 more cm left to the dotted line, that must be the answer to what I am looking for.

After I got all of the measurments needed to find the last side length, I did Pothagoras to find it out the last side length. For the last side I got a measurement of 8.5 cm.

Once I had all measurements possible, I calculated the surface area of the whole shape except the base face, and got an answer of 988u squared.

That is all I have to say! Thank you and have a great day!

(oooooh that rhymed)