Hello, prealgebrites.
Thank you to Sirus for starting us off. Below you will find a more complete list. Click to enlarge.
Be sure to review graphing relations and plotting points from chapter 1.
Post any questions to the blog. You should answer each other's questions. I may or may not answer them.
Good luck.
Monday, December 10, 2007
Thursday, December 6, 2007
Study Guide for the Final, a la Sirus
Study Guide for Wednesday’s Final
You Should Know How To:
Simplify algebraic expressions.
Solve basic equations, with variables and whole numbers. They will be one or two step equations.
Discern primes and “regular” or non primes.
Put numbers into Scientific Notation, small and large numbers
Create factor trees, and how to find the prime factorization of a number.
Find LCM’s and GCF’s.
Add, subtract, divide, and multiply fractions.
Convert fractions to decimals.
Simplify, and or change repeating decimals into fractions.
Simplify, or solve exponents.
You should know:
The division rules for 2,3,4,5,6,8,9,10
Remember the following properties.
Yeah, it's pretty small. If you need questions, just comment on this post.
You Should Know How To:
Simplify algebraic expressions.
Solve basic equations, with variables and whole numbers. They will be one or two step equations.
Discern primes and “regular” or non primes.
Put numbers into Scientific Notation, small and large numbers
Create factor trees, and how to find the prime factorization of a number.
Find LCM’s and GCF’s.
Add, subtract, divide, and multiply fractions.
Convert fractions to decimals.
Simplify, and or change repeating decimals into fractions.
Simplify, or solve exponents.
You should know:
The division rules for 2,3,4,5,6,8,9,10
Remember the following properties.
Yeah, it's pretty small. If you need questions, just comment on this post.
Tuesday, November 6, 2007
Hey guys its Gabriel with today’s lesson and POD
Treating each letter as a variable and treating each space as a multiplication symbol, write the following expression in expediential form:
Good golly miss Molly
The answer was: g2o4dl4y2is2m2 there were 2 G’s, 4 O’s, 1 D, 4 L’s, 2 Y’s, 1 I, 2 S’s, and 2 M’s.
Mr. A. asked if there were any H.W. questions. Lilly had one, it was:
And it was wrong. Sirus and Karol tried it and both got it wrong. You know what that means…. this problem is impossible (just kidding). Then Margo did it and did it right. The answer was
Mr. A. gave an example: it actually was because if the nominator has no value you put a 1 in it. By the way a nominator is the top part of a fraction.
Then Mr. A. asked us to do some problems.
1) A. Write 8 as a product of factors of 2. = 2x2x2
B. Write 8 as a product of powers of 2. = 21x21x21
C. Write 8 as a single power of 2. = 23
2) A. Write 16 as a product of factors of 14. =4x4
B. Write 16 as a product of factors of 22. =22x22
C. Write 16 as a single of a power of 2. =22
3) A. Write 27 as a product of factors of 3. = 3x3x3
B. Write 27 as the product of 2 with different powers of 3. =31x32
C. Write 27 as a single of powers of 3. = 33
4) A. Write 81 as a product of factors of 3. = 3x3x3x3
B. Write 81 as a product of factors of . = 31x32
C. Write 81 as a single of a power of 3. = 34
Since all of that was sort of difficult to read Mr. A. made a table which s the following:
8= 2x2x2 21x21x21 23
16= 4x4 22x22 24
27= 3x3x3 32x31 33
81= 3x3x3x3 31x32 34
amxan=? am-n
PRODUCT OF POWERS: am x an=am-n
If expressions with the same base are multiplied, then keep the base and add the powers.
Then we had to do some more math problems that Mr. A. gave us.
1) 23 x 25 = 23+5 = 28
2) 63x 63=66
3) (-9)2(-9)2= 94
4) 10t2 x 4t10 = 40t12
5) (4f8) (5f6) = 20f14
6) (8c2) (9c) = 72c3
None of the problems will be on the test on Wednesday. The key to these problems is you multiply the bases and add the exponents and leave your variable if you have one. Number 6 might seem a little confusing if you weren’t at class today. If there s no exponent, just pretend that there is a #1 in the empty space.
H.W. read pgs. 175-176 pg.178 13-23 odd, and 47, 57, 61.
REMINDER: The test on Wednesday consists of the following:
1) exponents
2) prime and composite numbers
3) divisibility rules
4) GCF of algebraic expressions
5) Simplifying fractions
6) And last but not least, writing in base 2 form(everyone’s favorite)
I nominate Lilly to be the next scribe!!!!!!!!!!!
Treating each letter as a variable and treating each space as a multiplication symbol, write the following expression in expediential form:
Good golly miss Molly
The answer was: g2o4dl4y2is2m2 there were 2 G’s, 4 O’s, 1 D, 4 L’s, 2 Y’s, 1 I, 2 S’s, and 2 M’s.
Mr. A. asked if there were any H.W. questions. Lilly had one, it was:
And it was wrong. Sirus and Karol tried it and both got it wrong. You know what that means…. this problem is impossible (just kidding). Then Margo did it and did it right. The answer was
Mr. A. gave an example: it actually was because if the nominator has no value you put a 1 in it. By the way a nominator is the top part of a fraction.
Then Mr. A. asked us to do some problems.
1) A. Write 8 as a product of factors of 2. = 2x2x2
B. Write 8 as a product of powers of 2. = 21x21x21
C. Write 8 as a single power of 2. = 23
2) A. Write 16 as a product of factors of 14. =4x4
B. Write 16 as a product of factors of 22. =22x22
C. Write 16 as a single of a power of 2. =22
3) A. Write 27 as a product of factors of 3. = 3x3x3
B. Write 27 as the product of 2 with different powers of 3. =31x32
C. Write 27 as a single of powers of 3. = 33
4) A. Write 81 as a product of factors of 3. = 3x3x3x3
B. Write 81 as a product of factors of . = 31x32
C. Write 81 as a single of a power of 3. = 34
Since all of that was sort of difficult to read Mr. A. made a table which s the following:
8= 2x2x2 21x21x21 23
16= 4x4 22x22 24
27= 3x3x3 32x31 33
81= 3x3x3x3 31x32 34
amxan=? am-n
PRODUCT OF POWERS: am x an=am-n
If expressions with the same base are multiplied, then keep the base and add the powers.
Then we had to do some more math problems that Mr. A. gave us.
1) 23 x 25 = 23+5 = 28
2) 63x 63=66
3) (-9)2(-9)2= 94
4) 10t2 x 4t10 = 40t12
5) (4f8) (5f6) = 20f14
6) (8c2) (9c) = 72c3
None of the problems will be on the test on Wednesday. The key to these problems is you multiply the bases and add the exponents and leave your variable if you have one. Number 6 might seem a little confusing if you weren’t at class today. If there s no exponent, just pretend that there is a #1 in the empty space.
H.W. read pgs. 175-176 pg.178 13-23 odd, and 47, 57, 61.
REMINDER: The test on Wednesday consists of the following:
1) exponents
2) prime and composite numbers
3) divisibility rules
4) GCF of algebraic expressions
5) Simplifying fractions
6) And last but not least, writing in base 2 form(everyone’s favorite)
I nominate Lilly to be the next scribe!!!!!!!!!!!
Monday, November 5, 2007
Sirus Needs Help
Hi guys. . . As the title says, it's me, Sirus. Actually, I didn't write down the homework, and Gaberiel is taking forever, so can anyone tell me what the homework is? Thanks in advance. . .
Thursday, November 1, 2007
what we did in pre-algebra nov. 1
HELLO,
Today in per-algebra class we did the following POD:
POD 11/1
An old lady wants to give her grandchildren her tresure. She has 135 gold coins and 105 esmeralds. She wants to give all them away.
a) what is the maximum number of grand children that can recieve an identical amount of esmeralds and coins?
b) how many esmeralds will each children get?
so the answer to the pod is the followingthe old lady can give it to 15 of her grandchildren for letter a, you can find this if you do a factor tree. and for leeter b she can give 7 esmeralds because if you divide15 into 105 you will get 7-
If you can try to find the GCF. We work on finding the GCF:
a. 24 and 60 and 18
b. 16x and 4x
c. 20xyy and 8xxy
We work on factoring Algecraic Expressions
5x + 45 = 5(x + 9)
- First to solve this problem you find the Gcf of 5 and 45.
The GCF= 5
-Second divide 5x and 45 by the GCF.
5x: 5= x and 45: 5= 9
-Then write the Gcf in the front of the parethesis
5( )
-Then put the solution of the division of the Gcf and the numerator
5( x + 9)
-This is the algebraic expression of 5x + 45=
Try to solve the following:
36 - 4x=
THEIR IS NO HOMEWORK FOR TONIGHT!!!!!!!!!!!!!!!!!!!!!!!!!!!!
THE NEW SCRIBE IS MARCO
Today in per-algebra class we did the following POD:
POD 11/1
An old lady wants to give her grandchildren her tresure. She has 135 gold coins and 105 esmeralds. She wants to give all them away.
a) what is the maximum number of grand children that can recieve an identical amount of esmeralds and coins?
b) how many esmeralds will each children get?
so the answer to the pod is the followingthe old lady can give it to 15 of her grandchildren for letter a, you can find this if you do a factor tree. and for leeter b she can give 7 esmeralds because if you divide15 into 105 you will get 7-
If you can try to find the GCF. We work on finding the GCF:
a. 24 and 60 and 18
b. 16x and 4x
c. 20xyy and 8xxy
We work on factoring Algecraic Expressions
5x + 45 = 5(x + 9)
- First to solve this problem you find the Gcf of 5 and 45.
The GCF= 5
-Second divide 5x and 45 by the GCF.
5x: 5= x and 45: 5= 9
-Then write the Gcf in the front of the parethesis
5( )
-Then put the solution of the division of the Gcf and the numerator
5( x + 9)
-This is the algebraic expression of 5x + 45=
Try to solve the following:
36 - 4x=
THEIR IS NO HOMEWORK FOR TONIGHT!!!!!!!!!!!!!!!!!!!!!!!!!!!!
THE NEW SCRIBE IS MARCO
Wednesday, October 31, 2007
Hey, It's the Halloween scribe post. ooooooooh.
Hey guys, it's me, Sirus again. I'm only doing this because Gaberiel isn't gonna be at his home until like 2:00 AM, so yeah. It's me.
Anyway, our POD.
Our POD was to write 11101 in base 10.
OK, so the way to figure this out follows.
To solve base 2, you go from right to left. So the last 1, in 11101, stands for 2 to the zeroeth, so 1. The zero is a place holder, and the 3rd digit, another 1 stands for 2 to the second, or 4. If you continue to do this, you will end up with 16+8+4+0+1, which equals 29. Since 29 is in base 10, 11101 in base 10 is 29.
After this, we practiced with some more binary, or base 2 numbers.
For example, 42 in base 2 is 101001.
40= 101000
53= 110101
55= 110111
Try finding 82 in base 2.
Next, we practiced turning base 2's into base 10's, with problems that we made up.
Gaberiels was 11110110, which turns into: 128+64+32+16+0+4+2+0, or 244. At least, that's what Gabe has on his paper of notes.
The last thing we learned was how to find the GCF, or Greatest Common Factor using factor trees. The steps were thus.
1) Find prime factors of each number.
2) Write the prime factors of each, writing out the exponents.
3) Circle all of the common factors.
4) Choose one row/set of the circled numbers.
5) Multiply the numbers.
One of our examples was 18 and 44.
The prime factorization of the two were:
Don't make fun of my picture, my wordperfect was on the fritz, so I just used paint.
Next, circle one 2 from each, as those are the only common factors. Next, multiply them.
You will find that the GCF is 2.
Another was 220 and 56, which if you will look:
The Homework was to read pages 164-166, and to do problems 17-29 (?) odd. Don't quote me on that. It might have been 17-31. For some reason, Gaberiel didn't write that down, but I'm not blaming him. Anyway, Mr. A said that he probably wouldn't collect the homework.
I pick (drum roll please) Karol to be the next scribe.
Anyway, our POD.
Our POD was to write 11101 in base 10.
OK, so the way to figure this out follows.
To solve base 2, you go from right to left. So the last 1, in 11101, stands for 2 to the zeroeth, so 1. The zero is a place holder, and the 3rd digit, another 1 stands for 2 to the second, or 4. If you continue to do this, you will end up with 16+8+4+0+1, which equals 29. Since 29 is in base 10, 11101 in base 10 is 29.
After this, we practiced with some more binary, or base 2 numbers.
For example, 42 in base 2 is 101001.
40= 101000
53= 110101
55= 110111
Try finding 82 in base 2.
Next, we practiced turning base 2's into base 10's, with problems that we made up.
Gaberiels was 11110110, which turns into: 128+64+32+16+0+4+2+0, or 244. At least, that's what Gabe has on his paper of notes.
The last thing we learned was how to find the GCF, or Greatest Common Factor using factor trees. The steps were thus.
1) Find prime factors of each number.
2) Write the prime factors of each, writing out the exponents.
3) Circle all of the common factors.
4) Choose one row/set of the circled numbers.
5) Multiply the numbers.
One of our examples was 18 and 44.
The prime factorization of the two were:
Don't make fun of my picture, my wordperfect was on the fritz, so I just used paint.
Next, circle one 2 from each, as those are the only common factors. Next, multiply them.
You will find that the GCF is 2.
Another was 220 and 56, which if you will look:
The Homework was to read pages 164-166, and to do problems 17-29 (?) odd. Don't quote me on that. It might have been 17-31. For some reason, Gaberiel didn't write that down, but I'm not blaming him. Anyway, Mr. A said that he probably wouldn't collect the homework.
I pick (drum roll please) Karol to be the next scribe.
Tuesday, October 30, 2007
Tuesday, October 30th
This is Austin, telling you what we did in class today.
The POD was the prime factorization of 224.
224
/\
2 112
/ /\
2 2 56
/ / /\
2 2 2 28
/ / / /\
2 2 2 4 7
/ / / /\ \
2x 2x2 x 2x2x7 = 2 to the 5th power x 7
2 more problems including prime factorization:
392= 2 to the 3rd power x seven to the 2nd power
-78= -2 x 3 x 13
then we did factoring expressions completly.
10x to the 2nd power = 2 times 5 times x times x times x
-48xy to the 3rd power= (-1) times 8 times 6 times x times y times y times y
-56ba to the 2nd power=(-1) times a times a b times 2 times 2 times 2 times 2 times 3
H.W is read Pg. 159-161. complete Pg. 161 #s 3, 13-39 odd.
we also learned that ato the 0 power is 1
1 example of expanded base 10 form is:
23=2 bundles of 10 and 3 bundles of 1
23=(2 x 10) + (3 x 1)
23=(2 x 10 to the 1st power)+ (3 x 10 to the 0 power)
1=10 to the 0 power 10=10 to the 1st power 100= 10 to the 2nd
346= (10 to the 2nd power x 3)+(10 to the 1st power x 4)+
(10 to the 0 power times 3)
4,003=(10 to the 3rd power x 4)+ (10 to the 0 power x 3)
computers use the base 2 form(bianary code).
1=on 0=off
Example:
21=10101
I choose Gabriel to be the next scribe.
The POD was the prime factorization of 224.
224
/\
2 112
/ /\
2 2 56
/ / /\
2 2 2 28
/ / / /\
2 2 2 4 7
/ / / /\ \
2x 2x2 x 2x2x7 = 2 to the 5th power x 7
2 more problems including prime factorization:
392= 2 to the 3rd power x seven to the 2nd power
-78= -2 x 3 x 13
then we did factoring expressions completly.
10x to the 2nd power = 2 times 5 times x times x times x
-48xy to the 3rd power= (-1) times 8 times 6 times x times y times y times y
-56ba to the 2nd power=(-1) times a times a b times 2 times 2 times 2 times 2 times 3
H.W is read Pg. 159-161. complete Pg. 161 #s 3, 13-39 odd.
we also learned that ato the 0 power is 1
1 example of expanded base 10 form is:
23=2 bundles of 10 and 3 bundles of 1
23=(2 x 10) + (3 x 1)
23=(2 x 10 to the 1st power)+ (3 x 10 to the 0 power)
1=10 to the 0 power 10=10 to the 1st power 100= 10 to the 2nd
346= (10 to the 2nd power x 3)+(10 to the 1st power x 4)+
(10 to the 0 power times 3)
4,003=(10 to the 3rd power x 4)+ (10 to the 0 power x 3)
computers use the base 2 form(bianary code).
1=on 0=off
Example:
21=10101
I choose Gabriel to be the next scribe.
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