this week we had the holidays..
umm, i did 39 of my 50 bonus point modules.
my computer was broken until wednesday so i did it thursday and friday.
& then a couple more on sunday. i hope not dong this doesn't make me fail. but whateva.
i don't remember what we were learning before we went into the holidays.
but on aleks i didn't know how to do the blue, purple, or yellow stuff.
so i went do some of the trig. stuff & what i had unlocked for the matrices thing.
but i still didn't know how to do anything else to get to my 50 modules.
so that isn't fair, JUS SAYin.
it's hot in this class.
Tuesday, March 15, 2011
Sunday, March 13, 2011
Nathan's blog
Yeah, so i didnt do my aleks and now I'm mad, so i might as well get some points for this. I'm probably just going to do a basic review of stuff we learned in class, and stuff in aleks.
So, I'm going to start off by reviewing matrices, which was very easy.
[2 4] [8 3] [10 7]
[7 5] + [9 12] = [16 17]
Now on to subtracting matrices.
[18 6] [5 7] [13 -1]
[15 3] - [14 9] = [ 1 -6]
Now, i will review how to find the slope of a line.
(4,3) (7,2)
2-4/7-3
The slope of the line is: -1/2
The slope of a line is 4/3. Find the slope of a parallel and perpendicular line, using this slope.
For the slope of a parallel line, it is the same as the slope of the original line. So the slope of this parallel line is: 4/3
When finding the slope of a perpendicular line, you take the negative reciporcal of the original slope. So the slope of this perpendicular line is: -3/4
That's it for this review.
So, I'm going to start off by reviewing matrices, which was very easy.
[2 4] [8 3] [10 7]
[7 5] + [9 12] = [16 17]
Now on to subtracting matrices.
[18 6] [5 7] [13 -1]
[15 3] - [14 9] = [ 1 -6]
Now, i will review how to find the slope of a line.
(4,3) (7,2)
2-4/7-3
The slope of the line is: -1/2
The slope of a line is 4/3. Find the slope of a parallel and perpendicular line, using this slope.
For the slope of a parallel line, it is the same as the slope of the original line. So the slope of this parallel line is: 4/3
When finding the slope of a perpendicular line, you take the negative reciporcal of the original slope. So the slope of this perpendicular line is: -3/4
That's it for this review.
Friday, March 11, 2011
Taylor's 14th(I think XD) Blog Review
REVIEW TIME!!!!
This is the review on my notes from Ch. 13.2.
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=4 tn=8tn-1+7
t1=8 (4)+7= 39 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 8 (39) + 7 = 319
t3= 8 (319) + 7= 2559
t4= 8 (2559) + 7 = 20479
t5= 8 (20479) + 7 = 163839
So your answers will be 2559, 20479, and 163839.
Now let's try finding the recursive definition for 8, 12, 16, 20...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 4 in between each number your answer or equation will be tn-1+4.
This is the review on the rest of my notes on Ch. 13.2. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
This is the review on my notes from Ch. 13.2.
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=4 tn=8tn-1+7
t1=8 (4)+7= 39 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 8 (39) + 7 = 319
t3= 8 (319) + 7= 2559
t4= 8 (2559) + 7 = 20479
t5= 8 (20479) + 7 = 163839
So your answers will be 2559, 20479, and 163839.
Now let's try finding the recursive definition for 8, 12, 16, 20...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 4 in between each number your answer or equation will be tn-1+4.
This is the review on the rest of my notes on Ch. 13.2. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
Sunday, March 6, 2011
Nathan's blog
So, we just took our exams on chapter 5 which dealt with logs and interest, and a few more formulas. I thought I understood the information, but apparently not. I just finished the 14 aleks modules that we needed done by today, and then tommorrow i will start on the 50 modules for extra credit.
So, now i will just review different stuff throughout chapter 5.
Rule of 72
72/r%=how long it takes to double
How long will it take to double your money at a 12% interest?
72/12=6
We learned a formula for interest:
A(t)=Ao b^t/k
Ex. A bank advertises that if you open a savings account, you can double your money in twelve years. Find out how much money you will have after 7 years, if you invest $1,000 into your account.
A(t)=1000(2)^7/12=$1498.31
I'm very happy to get out of chapter 5, hopefully i will understood the information better in the 4th nine weeks, because that will probably be the toughest part of the year in math.
So, now i will just review different stuff throughout chapter 5.
Rule of 72
72/r%=how long it takes to double
How long will it take to double your money at a 12% interest?
72/12=6
We learned a formula for interest:
A(t)=Ao b^t/k
Ex. A bank advertises that if you open a savings account, you can double your money in twelve years. Find out how much money you will have after 7 years, if you invest $1,000 into your account.
A(t)=1000(2)^7/12=$1498.31
I'm very happy to get out of chapter 5, hopefully i will understood the information better in the 4th nine weeks, because that will probably be the toughest part of the year in math.
Nicala's Post
okay we had our exams and we learned the rest of chapter five which was about exponents which we learned in aleks and logs which we should have learned in algebra 2 but i did not understand it then and do not understand now. okay we took our exam thursday which i failed horrible and we started doing fourteen aleks friday and will finish them this weekend and tomorrow we start the fifty bonus aleks for the extra fifty points in class which i need. one of things i learned on aleks is how to do translation of a vector.
Example Problem
A vector with an initial point of (-6,-2) and terminal point of (-5,3) is translated so that its initial point is at the origin. Find its new terminal point.
To find the terminal point you have subtract them from each other.
you subtract -5 minus -6. since you are subtracting and the six is negative the six becomes positive which gives you 1 as your first new terminal point.
next you subtract 3 minus -2 since you are subtracting and the two is the two becomes positive
which gives you 5 as your second new terminal point.
the answer is (1,5)
Example Problem
A vector with an initial point of (-6,-2) and terminal point of (-5,3) is translated so that its initial point is at the origin. Find its new terminal point.
To find the terminal point you have subtract them from each other.
you subtract -5 minus -6. since you are subtracting and the six is negative the six becomes positive which gives you 1 as your first new terminal point.
next you subtract 3 minus -2 since you are subtracting and the two is the two becomes positive
which gives you 5 as your second new terminal point.
the answer is (1,5)
Friday, March 4, 2011
Taylor's 14th(I think XD) Blog Review
REVIEW TIME!!!!
This is the review on the rest of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 3 x 2^ n-1
t1=3 x 2^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=3 x 2^ 2-1
t3=3 x 2^ 3-1
t4=3 x 2^ 4-1
t1= 3 Then you have your answers.
t2= 6
t3= 12
t4= 24
9,36,144,576.... find the formula for the nth term.
9,36,144,576.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 4.
tn= t1 x r ^n-1Then write out your equation.
tn= 9 x4 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
This is the review on the rest of my notes on Ch. 13.1. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
This is the review on the rest of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 3 x 2^ n-1
t1=3 x 2^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=3 x 2^ 2-1
t3=3 x 2^ 3-1
t4=3 x 2^ 4-1
t1= 3 Then you have your answers.
t2= 6
t3= 12
t4= 24
9,36,144,576.... find the formula for the nth term.
9,36,144,576.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 4.
tn= t1 x r ^n-1Then write out your equation.
tn= 9 x4 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
This is the review on the rest of my notes on Ch. 13.1. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
Charlie.
This week we had exams.
But anyways, in aleks i ended up having to take the stupid assessment thingy & got knocked down to like 70 in my blue thingy..
It's friday, in first hour.
& i think today in math class we'll be coming to the lab or A1 to do 7 alek modules.
but not the library because the book fair is this week, so yea.
then, on monday we could start on the extra credit thingy.
extra credit this nine weeks is 50 aleks modules.
which, idk how i'm goin to do that because MY COMPUTER'S BROkEN!
but, i guess i'll figure something out because i would like a really good grade this time.
... This week we learned 2 (or 3?) new sections in chapter (i think) 5.
then we pretty much reviewed for the rest of the week and took our exam yesterday (thursday).
when i was doing the practice chapter test in the text book i knew what i was doing,
but when i actually took the exam; i was like "WhAt?!?"
so, (basically) i do believe that i FAILED(:
example of something:
log2~x = 4
2^4 = x
x = 16
But anyways, in aleks i ended up having to take the stupid assessment thingy & got knocked down to like 70 in my blue thingy..
It's friday, in first hour.
& i think today in math class we'll be coming to the lab or A1 to do 7 alek modules.
but not the library because the book fair is this week, so yea.
then, on monday we could start on the extra credit thingy.
extra credit this nine weeks is 50 aleks modules.
which, idk how i'm goin to do that because MY COMPUTER'S BROkEN!
but, i guess i'll figure something out because i would like a really good grade this time.
... This week we learned 2 (or 3?) new sections in chapter (i think) 5.
then we pretty much reviewed for the rest of the week and took our exam yesterday (thursday).
when i was doing the practice chapter test in the text book i knew what i was doing,
but when i actually took the exam; i was like "WhAt?!?"
so, (basically) i do believe that i FAILED(:
example of something:
log2~x = 4
2^4 = x
x = 16
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