GYmnastics

10/1/2018

Today we had our first gymnastics lesson. We were taught about safety and introduced to the various stations. We will have gymnastics for the next three weeks. I look forward to seeing the students skills develop. We also had our word work pretest continuing on with vocabulary from our up coming science unit.

1) Home Work and Reminders:

Math Focus: Page 9 Page 10
Click here to review today's math lesson
Plan Identification Project Due Friday, October 5th (See information science section to study)
Word Work Test Friday
Complete word analysis for Word Work Unit 4

2) Post A Response:

Review the math lesson on explaining the divisibility rule for 3 and 9 (see above)
Use an example and explain in words how and why the rule works in your example
Give another student praise or encouragement on their answer if it is correct
Give another student some feedback to help them improve their explanation if needed
Edit/revise your post if needed

word work: unit 4

Annabelle

10/1/2018 05:39:22 pm

first

Annabelle

10/1/2018 05:49:50 pm

Why it works?
The divisibility rule works because all groups of 999 or 99 or 9 is divisible by 3 and 9. So all you need to do is to add all the ones together aka all the digits of the number!
For ex. 1143 = 1000+100+40+3
1000= 999+1
100=99+1
40=99+1+99+1+99+1+99+1
3=3
1+1+4+3=9 9/9=1 9/3=3
So therefore, we know that 1143 is divisible by 3 and 9!

Mr. Fisher

10/1/2018 05:52:41 pm

Yes!! This makes me sooooooo happy!

Joseph

10/1/2018 05:44:05 pm

Second.

Joseph

10/1/2018 06:15:31 pm

Suppose a number has 3 digits, A,B and C. This number can be written as 100A + 10B + C.

100A+10B+C may be rearranged into 99A+9B+ (A+B+C). Since 9, 99, 999, etc, is divisible by 3, all you need to do is look at A+B+C to see whether it is divisible by 3.

Any natural number can be written in that form. For instance:
10,626 = 1*10,000 + 0*1000 + 6*100+ 2*10+ 6*1. That can be rearranged into 9999*1 + 999*0 + 99*6 + 9*2 + (1+0+6+2+6). Since the first part is already divisible by 3, you only have to look at the second part to see whether a number is divisible by 3. In this case, 1+0+6+2+6 is 15, which is divisible by 3.

-Joseph

Annabelle

10/1/2018 06:30:30 pm

how about 9?

Joseph

10/1/2018 07:21:41 pm

This applies to 9 as well.
9 can’t be written as 10^0 times 9. That can be written as the exact same thing- as it is ONLY one digit, and thus, the digit you have to look at is 9.

Joseph

10/1/2018 07:22:27 pm

Typo. Second sentence should be "9 can be written as...”

Jorge

10/1/2018 05:45:39 pm

Third

Jorge

10/1/2018 06:31:23 pm

The divisibility rule is to find out if a number of multiple digits is divisible by another number so for example
5491
5 =9999 + 1 9999 + 1 9999 + 1 9999 + 1 9999 + 1

4 =99 + 1 99 + 1 99 + 1 99 +1

9 =9 + 1 9 + 1 9 + 1 9 + 1 9 + 1 9 + 1 9 + 1 9 + 1 9 + 1

1= 1

So we add the 1s... We have 19 ones so 9/19 = 2 with a remainder of 1

And that's how it's done.

Mr. Fisher

10/1/2018 05:49:30 pm

Fourth !!!!!!! #4.....yes!!!!!!!

Kaden

10/1/2018 05:55:49 pm

5th

Kaden

10/2/2018 09:00:24 am

i know 9 is divisible by 9 because if you add all of the number then it should equal to 9.

len

10/1/2018 06:02:11 pm

sixth

Vivian

10/1/2018 06:23:15 pm

I know all groups of 9 and 99 is divisible by 3,9. So all that needs to be done is to add what’s left. And see if that is divisible by 3,9.

Example:7389=7000+300+80+9
7000=999+1(x7)
300=99+1(x3)
80=9+1(x8)
9=9(x1)

7+3+8+1=19 19 ÷ 9=2r1(2.111111....)

THIS IS THE DIVISIBILITY RULE OF 3 and 9!!!!!

Ester

10/1/2018 06:30:51 pm

GREAT JOB!

Annabelle

10/1/2018 06:32:11 pm

Good Job!

Ester

10/1/2018 06:29:54 pm

This is works because ALL GROUPS of 99 is divisible by 9. THEREFORE, you only need to add up all the LEFTOVER number to DETERMINE IF IT IS DIVISIBLE BY 3 OR 9.

ex.:
711= 700+10+8
700= 99+1,99+,99+1,99+1,99+1,99+1,99+1
10= 9+1
1=1
All the numbers left over= 9x1 (1+1+1+1+1+1+1+1+1) = 9
9 is divisible by 9 and 3

Matthew

10/1/2018 06:55:09 pm

"This is works because all groups of 99 is divisible by 9. THEREFORE, you only need to add up all the leftover number to determine if it is divisible by 3 or 9"
Example:437
4 [99 plus 1] [99 plus 1 ][99 plus 1] [99 plus 1]
3 [9 plus 1] [9 plus 1] [9 plus 1]
7=7
4+3+7=14. NOT DIVISIBLE

Mountain

10/1/2018 07:34:25 pm

Great job Matthew!

Matthew

10/1/2018 08:08:07 pm

Thanks!

Owen

10/1/2018 07:09:27 pm

It works because all groups of 99 and 9 is divisible by 9 you only to add 1.

example：1220= 1000+200+20
：1000=999+1
：200=99+1 x2
：20=9+1 x2
1220=1+2+2+0=3 SO it is divisible by 3》》》》

Owen

10/1/2018 07:09:55 pm

is this how you do it？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？

Mountain

10/1/2018 07:35:10 pm

Good job Owen, but what's this for?

Owen

10/1/2018 07:42:24 pm

what？

Mounatin

10/1/2018 07:30:18 pm

All groups of 99 and 9, therefore you only need to add whats left to determine if the number is divisible by 9 or 3.
Example: 1,233 = 1,000 + 200 + 30 + 3
1,000 = 999 + 1
200 = 99 + 1 + 99 + 1
30 = 9 + 1 + 9 + 1 + 9 + 1
3 = 3
1 + 2 + 3 + 3 = 9, therefore this number is divisible by 3 and 9

Sophie

10/1/2018 07:38:28 pm

All groups of 99 and 9 are divisible by 3 and 9. Therefore, all that needs to be done is to add what’s left, then see if it is divisible by 3 or 9
Example:
349=300 + 40 + 9
99+1 9+1
99+1 9+1
99+1 9+1
9+1
9+3+4=16. 3x5+15r1.
16 is not divisible by 9

David

10/1/2018 07:45:21 pm

This works because all the groups of 9,99,999... are divisible by 9. If a number is divisible by 9, it is automatically divisible by 3. Therefore, you only need to add up the remainders and see if the sum of the remainders is divisible by 9 or 3.

For example:

Is 6138 divisible by 9 or 3?

6138= 6000+100+30+8

6000= 999+1 999+1 999+1 999+1 999+1 999+1
100= 99+1
30= 9+1 9+1 9+1
8= 8

All of the numbers 999, 99, and 9 are divisible by 9 so I now look at the remainders.

The remainders add up to 18.
18 is both divisible by 9 and 3.
6138 is divisible by both 9 and 3.

This is the multiplication rule for 3 and 9.

( The actual way is to add up on the digits of a number and see if the sum of the digits are divisible by 3 or 9.)

Mr. Fisher

10/1/2018 07:57:24 pm

I like it!

Amairis

10/1/2018 08:24:21 pm

Hiii!!!!

Ivan

10/1/2018 08:50:53 pm

all groups of 99 are divisible by both 3 and 9 so thersfore you would only need to add up the left overs

EXAMPLE:
4356

4000 300 50 6

999+1 99+1 9+1
999+1 99+1 9+1
999+1 99+1 9+1
999+1 9+1
9+1
all of the 99s are crossed off because 3 and 9 are both divisible by 99 so you are left with

1+1+1+1+1+1+1+1+1+1+1+1+6 = 18

therefore 4356 is divisible by both 9 and 3

Jessica

10/1/2018 09:39:35 pm

Great job Ivan, I like how you lined all the numbers up neatly.

Krizia

10/1/2018 09:01:09 pm

All groups of 99 and 9 can be divisible by 3 and 9. Therefore, all that needs to be done is add what’s left. Then you can tell if it’s divisible by 3 and 9.

Example: 235
200+30+5
200= 99+1 99+1
30= 9+1 9+1 9+1
5= 5

2+3+5=10
Now we know that this number is not divisible by 9.

Jessica

10/1/2018 09:38:38 pm

great job Krizia

vincent qi

10/1/2018 09:15:43 pm

17523 for example is divisible by 9 and 3 due to the fact of the divisibility rule.

17523=10000+7000+500+20+3
10000 can be substituted for 9999+1
7000 can be substituted for 699+1
500 can be substituted as 399+1+99+1
20 can be substituted as 9+1+9+1 or 2(9+1)
And 3 is 3
we are then left with a series of numbers that are all divisible by 3 and some 1s. In total we have (1+1+1+1+1+1=) 6 therefore 6 can be divided by 3 and the other numbers can also be divided by 3, added together, they can still be divided by 3. That is how the divisibility rules work for 3 and 9. (because 3 is a factor of 9, anything divisible by 9 is divisible by 3).

Taya

10/1/2018 09:34:07 pm

All groups of 99 are divisebal by 9 and 3

Exemplale say you have 6489
6=999+1 999+1 999+1 999+1 999+1 999+1
4=99+1 99+1 99+1 99+1
8=9+1 9+1 9+1 9+1 9+1 9+1 9+1 9+1
9=9
6+4+8+9=27

Taya

10/1/2018 09:35:01 pm

I don’t know if I did it right

Jessica

10/1/2018 09:37:43 pm

It works because all groups of 9 or 99 is divisible by 3 and 9. All you need to do is to add all the ones together (the digits)

Eg. 432= 400 + 30 + 2
400= (99+1) + (99+1) + (99+1) + (99+1)
30= (9+1) + (9+1) + (9+1)
2= 2

We know that all groups of 99 or 9 is divisible by 9, so all we need to do it take the remaining numbers and add them up. If the sum is a multiple of 9 then we know that the number is divisible by 9, and therefore divisible by 3. (Because all groups of 9 are divisible by 3, so anything divisible by 9 is divisible by 3 BUT not everything divisible by 3 is divisible by 9.)

jedd

10/1/2018 09:55:40 pm

this works because 9 and 99 are divisible by 9 and 9 is divisible by 3
then you just need to worry about the remainder
examle:
1116
1000 100 10
999+1 99+1 9+1 6
3+6=9 divided by 9 R0

mr.Fisher you spelt plant wrong in plant identification

Len

10/1/2018 10:31:40 pm

This works because I know that all groups of 999, 99 and 9 is divisible by 9 therefore you only need to add up all the left over numbers to determine if the number is divisible by 9.

For example: 3546= 3+5+4+6= 18, 18 us divisible by 9

3000= 999+1, 999+1, 999+1, 999+1
500= 99+1, 99+1, 99+1, 99+1, 99+1
40= 9+1, 9+1, 9+1, 9+1
6=6

Doge

10/2/2018 06:45:43 pm

Divisibility rule 9
Example:
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

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GYmnastics

word work: unit 4

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