This week we learned a section in chapter 5 which deals with rational exponents. This section is a easy and not that hard at all.
Formulas:
A(t)=A0b^(t/k)
-A0 is starting point
-b is halfing, doubling, ect
-t is time and k is time it takes to get to b
Examples:
3rd root of square root of x= x^1/3
(16/100)^1/2= square root of 16/square root of 100 which can be simplified to 2/5
27^2x=3^2
(3^3)^2x=3^2
3^6x=3^2
6x=2
x=1/3
Basically if you know your rules and the formula this section will be easy..
Sunday, February 27, 2011
chapter 5
This week we learned things from chapter 5 and completed a lot of aleks modules. 5 was about exponents and a couple formulas. Unfortunately, exponents are not the easiest thing for me which is why I think I did so much worse in algebra II. Since I’m not very good at those, I’ll do a formula example.
The first formula we did was the easiest for me. It is:
A(t) = Ao (1 + r)^t
This formula is used with word problems when given a percent, time and rate.
Ao= your starting number
R= the rate
T= time
After you read the problem, you plug in the numbers to the formula, type it in your calculator, and it gives you the answer. (percents must be changed to decimals)
Ex: Suppose you have $1,000 dollars invested at a 7% interest rate. How much money will you have after 2 years?
A(t) = $1,000 (1 + .07)^2
^^ plug that into your calculator and you get $1144.90 as your answer.
The first formula we did was the easiest for me. It is:
A(t) = Ao (1 + r)^t
This formula is used with word problems when given a percent, time and rate.
Ao= your starting number
R= the rate
T= time
After you read the problem, you plug in the numbers to the formula, type it in your calculator, and it gives you the answer. (percents must be changed to decimals)
Ex: Suppose you have $1,000 dollars invested at a 7% interest rate. How much money will you have after 2 years?
A(t) = $1,000 (1 + .07)^2
^^ plug that into your calculator and you get $1144.90 as your answer.
Chapter 5
This week in advanced math we had to complete seven aleks modules and this weekend we have to do fourteen. I finished both so I’m super glad. Mrs. Robinson wasn’t here for a few days but we did manage to start chapter five. We were also supposed to get packets this week for our exam this Friday. However, considering the circumstances that happened this week Mrs. Robinson changed the exam to just chapter five. Chapter five is about exponents. We covered a few sections in five and it is pretty easy stuff. We learn how to solve exponents and we used 3 different formulas. Let’s do some example problems.
Solve the following problems:
Example 1:
7-2 = 1/49
Example 2:
(-7)-2 = -1/49
Example 3:
When given a chart you must pick one of the following formulas:
Read the problem, plug in the numbers, and then into your calculator and get an answer.
A(t) = Ao (1+r)2
A(t) = Ao b t/k
P(t) = Po ert
Solve the following problems:
Example 1:
7-2 = 1/49
Example 2:
(-7)-2 = -1/49
Example 3:
When given a chart you must pick one of the following formulas:
Read the problem, plug in the numbers, and then into your calculator and get an answer.
A(t) = Ao (1+r)2
A(t) = Ao b t/k
P(t) = Po ert
Rational Exponents
This week in advanced math I think it is chapter 5 which we are working on. One of the sections we are doing is called rational exponents. This is not to difficult I will show you how to do some of these kinds of problems.
FORMULAS:
A(t)=A0b^(t/k)
A0 is starting point
b is halfing, doubling, ect
t is time and k is time it takes to get to b
Exponents that involve roots.
EXAMPLE:
5th root of (3)^3
3^3/5
(4/16)^1/2
2/4 which is equal to 1/2
*This is the easiest way to explain rational exponents.
FORMULAS:
A(t)=A0b^(t/k)
A0 is starting point
b is halfing, doubling, ect
t is time and k is time it takes to get to b
Exponents that involve roots.
EXAMPLE:
5th root of (3)^3
3^3/5
(4/16)^1/2
2/4 which is equal to 1/2
*This is the easiest way to explain rational exponents.
this past week was the worst week of school ever. we learned about rational exponents. these are very easy and we've seen this in aleks. everything we did this week was referred to aleks. so here are a few examples of stuff.
EXAMPLE:
6square root of x^3--------x^3/6
FORMULAS:
a(t)=a0(1+r)^t
0---is your starting point
r---is your decimal
t---is your time
*when given a word problems you can easily plug it into the formula. once you have it plugged into your formula you can then put it in your calculator to get your final answer. It's really simple so i hope we have a good bit of these on our 9 weeks exam friday :)
i've still been trying to do aleks but it keeps kicking me off my computer. i am learning with using aleks.
EXAMPLE:
6square root of x^3--------x^3/6
FORMULAS:
a(t)=a0(1+r)^t
0---is your starting point
r---is your decimal
t---is your time
*when given a word problems you can easily plug it into the formula. once you have it plugged into your formula you can then put it in your calculator to get your final answer. It's really simple so i hope we have a good bit of these on our 9 weeks exam friday :)
i've still been trying to do aleks but it keeps kicking me off my computer. i am learning with using aleks.
Exponents
This was a hectic week in school and advanced math. We learned about all kinds of exponents and how to change them from exponents to radicals and vise versa. We did a lot in Aleks, and we had a lot of homework this week. We also learned about rational exponents, which are exponents that involve roots. This was pretty easy with just the numbers. We we inserted letters it took me a little while to get the hang of it, but I figured it out. I enjoyed Rational Exponents better than the T1 formula so here you go.
Rational Exponents
Ex. (49/81)^1/2
1) (49^1/2)/(81^1/2)
2)(square root of 49/ square root of 81)
3)7/9
5-4
Almost forgot about this! This week in Advanced Math we learned about exponents and formulas involving percent, rate, and time. Section 5-4 involved one formula – P(t)=Pe^r*t
P – starting amount
e – on calculator (e is approximately 2.7)
r – rate as a decimal
t – time
EXAMPLE 1: Suppose $2,000 is invested at 5% interest compounded continuously. How much money do you have after 3 years?
A). Using the formula, plug in the information given.
P=2,000(starting amount)e^(0.5*3)
B). By plugging all of that into your calculator, you get $8963.38 as your final answer.
EXAMPLE 2: Which has a larger value? e^square root of 3 or 3^e
A). The answer to this problem can be achieved by plugging the two into your calculator.
B). e^square root of 3 = 5.65 and 3^e = 19.81
C). Therefore, the first one has a larger value.
This is a little short but my computer is installing new updates and it’s counting down to shutdown!
Almost forgot about this! This week in Advanced Math we learned about exponents and formulas involving percent, rate, and time. Section 5-4 involved one formula – P(t)=Pe^r*t
P – starting amount
e – on calculator (e is approximately 2.7)
r – rate as a decimal
t – time
EXAMPLE 1: Suppose $2,000 is invested at 5% interest compounded continuously. How much money do you have after 3 years?
A). Using the formula, plug in the information given.
P=2,000(starting amount)e^(0.5*3)
B). By plugging all of that into your calculator, you get $8963.38 as your final answer.
EXAMPLE 2: Which has a larger value? e^square root of 3 or 3^e
A). The answer to this problem can be achieved by plugging the two into your calculator.
B). e^square root of 3 = 5.65 and 3^e = 19.81
C). Therefore, the first one has a larger value.
This is a little short but my computer is installing new updates and it’s counting down to shutdown!
Sunday, February 20, 2011
This was a wild week in math and school in general. From matrices to carnival ball, we experienced it all this week. Now it is time for my least favorite part of the night. The part where I have to blog late at night, and come to the realization that the weekend is over and school is tomorrow. But anyways heres your blog.
Matrices
This week in advanced math we learned how to do matrices. It is a long tedious process but it isnt too hard.
Ex.
x y z
2 3 4
3 2 1
x[3*1 + 4*2] + y[2*1 + 3*4] + z[2*2+3*3] = x[11]+y[14]+[13]= 38
blog.blog.blog.
hellooo everyonee:)
well this week everyone has really been busy with the carnival ball and what not. But despite all the hectic over that i must admit that the week was pretty productive for my advanced math and physics classes anyway..but anywho we learned a lot vectors again. When we were'nt learnning about vectors, we were working on our ALEKS which is gonna be done pretty soon.
we learned how to tell if vectors were parallel or perpendicular..you can do this by doing dot product. if the answer is equal to zero, then vectors are perpendicular.
ex. ( 2,3) (0,0)
(2*0)+(0*3)
=0+0
=0
therefore, the two vectors are perpendicular :)
in order for two vectors to be parallel they must be proportional. Here is an example of two vectors that are proportional, and parallel.
ex. x(1,2) y(2,3) z(3,4)
2/1=3/2=4/3
2=3/2=4/3
so noo, these vectors are not parallel which is too bad :(
well this week everyone has really been busy with the carnival ball and what not. But despite all the hectic over that i must admit that the week was pretty productive for my advanced math and physics classes anyway..but anywho we learned a lot vectors again. When we were'nt learnning about vectors, we were working on our ALEKS which is gonna be done pretty soon.
we learned how to tell if vectors were parallel or perpendicular..you can do this by doing dot product. if the answer is equal to zero, then vectors are perpendicular.
ex. ( 2,3) (0,0)
(2*0)+(0*3)
=0+0
=0
therefore, the two vectors are perpendicular :)
in order for two vectors to be parallel they must be proportional. Here is an example of two vectors that are proportional, and parallel.
ex. x(1,2) y(2,3) z(3,4)
2/1=3/2=4/3
2=3/2=4/3
so noo, these vectors are not parallel which is too bad :(
Alright this week we were really crazy. There was carnival ball and stuff. We took a test on Thursday and Friday we didn’t have math class and a bad thing is I forgot my whole binder. If I do it on something easy I’m sorry and hope you forgive me.
Examples:
Matrices:
2[ 1 2 4] + 3{3 2 5]
{2 4 8} + { 9 6 15}= [ 11 10 23}
Named the matrices
{ 1 3 4 5}
{ 2 4 8 9} = 2x4
find AB
A(1 4 8) B(5 4 7)= (4, 0, -1)
point 8,-2, 3 and 4, 0, 7
Find center = (6,-1,5)
Radius square root (4-8)^2+(0+2)^2(7-3)^2=16+4+16 = square root of 36= 6
Equation = (x-6)^2+(y+1)^2+(2-5)^2=9^2
Examples:
Matrices:
2[ 1 2 4] + 3{3 2 5]
{2 4 8} + { 9 6 15}= [ 11 10 23}
Named the matrices
{ 1 3 4 5}
{ 2 4 8 9} = 2x4
find AB
A(1 4 8) B(5 4 7)= (4, 0, -1)
point 8,-2, 3 and 4, 0, 7
Find center = (6,-1,5)
Radius square root (4-8)^2+(0+2)^2(7-3)^2=16+4+16 = square root of 36= 6
Equation = (x-6)^2+(y+1)^2+(2-5)^2=9^2
This week we learned a few new sections, had to complete a total of 70 modules on aleks, and took our Chapter 12 test. The next week our whole blue pie will have to be complete. In the section below I will explain our notes on 3D. This Chapter is mostly formulas and is pretty easy to remember if you study it.
Formulas you need to know for this section are below:
AB = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2
Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)
Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2
Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)
Vector addition, magnitude, dot product, etc.are just a little different but follow the same formulas as earlier sections. They just have an extra variable in the formula unlike before.
Example
Find the center and radius Point A (8,4,2) Point B (2, 8,6)
Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)
Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 6+ 6 + 6= square root 16/2 = square root 8
Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 8
Formulas you need to know for this section are below:
AB = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2
Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)
Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2
Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)
Vector addition, magnitude, dot product, etc.are just a little different but follow the same formulas as earlier sections. They just have an extra variable in the formula unlike before.
Example
Find the center and radius Point A (8,4,2) Point B (2, 8,6)
Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)
Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 6+ 6 + 6= square root 16/2 = square root 8
Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 8
Alright soooo, im glad carnival ball is over. But now i'm going to fail this class. ALEKS is not my friend anymore. I work too slow and i havent met the latest deadline which will cause my grade to go wayyyyyyy down. The assessment test sucksssssssssssss.
glad ive said that now on to some math.
Finding whether something is parallel or perpendicular is my favorite.
EXAMPLE:
j=(3,4,2) k=(15,8,4)
PARALLEL:
15/3=5
8/4=2
4/2=2
PERPENDICULAR:
3(15)+4(8)-2(4)
**in order for it to be perpendicular it must equal 0.
****therefore its neither.
The AB thing is easy too.
A(2,3,1) B(4,2,6)
4-2=2
2-3=-1
6-1=5
*your answer is (2,-1,5)
my computer keeps freezing so im going to post now.
glad ive said that now on to some math.
Finding whether something is parallel or perpendicular is my favorite.
EXAMPLE:
j=(3,4,2) k=(15,8,4)
PARALLEL:
15/3=5
8/4=2
4/2=2
PERPENDICULAR:
3(15)+4(8)-2(4)
**in order for it to be perpendicular it must equal 0.
****therefore its neither.
The AB thing is easy too.
A(2,3,1) B(4,2,6)
4-2=2
2-3=-1
6-1=5
*your answer is (2,-1,5)
my computer keeps freezing so im going to post now.
Hey guys so this blog is going to suck this week because I have had a super, busy, exhausting weekend with carnival ball, and we had a field trip and we did not have class on Friday, and I forgot to bring my notebook home so I am going to have to go off of some super easy stuff I remember so forgive me. Let’s see matrices are pretty easy so let’s just do some random examples cause I did awesome on my on this entire chapter! Yay!
Random example of matrices:
Solve the matrices:
2 [ 3 4 5 ] = [ 6 8 10 ]
[ 7 8 ] + [ 9 10 ] = [ 16 18 ]
Name the matrices:
[ 2 4 5 6 9 ] is a 1 x 5
[ 9 10 15 ]
[ 1 12 20 ] is a 2 x 3
Random example of matrices:
Solve the matrices:
2 [ 3 4 5 ] = [ 6 8 10 ]
[ 7 8 ] + [ 9 10 ] = [ 16 18 ]
Name the matrices:
[ 2 4 5 6 9 ] is a 1 x 5
[ 9 10 15 ]
[ 1 12 20 ] is a 2 x 3
Vectors
This week in advanced math we have just finished chapter 12. Chapter 12 was about vectors. We took a test on vectors and I thought that the test was fairly easy. The hardest part was the 4x4 matrix which wasnt really that hard; just time consuming. Today I am going to teach you how to add and subtract a vector. This is probably one of the easiest things you will learn to do.
DIRECTIONS:
1. You should make sure you have a vector, u and v are always vectors.
2. Then you should add x1 to x2 and you will come out with a new vector.
3. You should do the same thing for subtraction just simply subtract the vectors.
EXAMPLES:
u=<4,5> v=<3,4>
Addition:
Find u+v
<4,5>+<3,4>
answer:<7,9>
Subtraction:
Find u-v
<4,5>-<3,4>
answer:<1,1>
*This is the easiest thing you will learn to do with vectors from my point of view.
*Just remember add the x's and add the y's and the same rules apply for subtraction.
DIRECTIONS:
1. You should make sure you have a vector, u and v are always vectors.
2. Then you should add x1 to x2 and you will come out with a new vector.
3. You should do the same thing for subtraction just simply subtract the vectors.
EXAMPLES:
u=<4,5> v=<3,4>
Addition:
Find u+v
<4,5>+<3,4>
answer:<7,9>
Subtraction:
Find u-v
<4,5>-<3,4>
answer:<1,1>
*This is the easiest thing you will learn to do with vectors from my point of view.
*Just remember add the x's and add the y's and the same rules apply for subtraction.
3x3
This past week we learned how to do 3x3's and 4x4's. I will show examples of the 3x3 because the 4x4's take FOREVER and so much paper. However, they are pretty much the same concept so if you are able to do one, you can do the other. 3x3's are very simple once you are taught the correct way because it is all simple math. So, here we go:
Example 1: (I won't have the brackets around them because I don't know how to do that)
6 2 3
1 2 4
0 1 1
1. you want to pick the row with the smallest numbers and cross it out.
2. cross out the first column, so you are left with:
0 * 2 3
2 4
3. Now cross out the second column, and so on ( alternate the sigsn of subtraction and addition.)
0 * 2 3 + 1 * 6 3 - 1 * 6 2
2 4 1 4 1 2
4. Now just multiply your matrices.
8-5= 3 *0 = 0
24-3 = 21 * 1 = 21
12-2= 10 * 1 = 10
0 +21 -10= 11
11 is your answer!
Example 1: (I won't have the brackets around them because I don't know how to do that)
6 2 3
1 2 4
0 1 1
1. you want to pick the row with the smallest numbers and cross it out.
2. cross out the first column, so you are left with:
0 * 2 3
2 4
3. Now cross out the second column, and so on ( alternate the sigsn of subtraction and addition.)
0 * 2 3 + 1 * 6 3 - 1 * 6 2
2 4 1 4 1 2
4. Now just multiply your matrices.
8-5= 3 *0 = 0
24-3 = 21 * 1 = 21
12-2= 10 * 1 = 10
0 +21 -10= 11
11 is your answer!
Saturday, February 19, 2011
3-D
The formula to find midpoint is (x1+x2/2, y1+y2/2, z1+z2/2)
IABI = square root (x2-x1)squared +(y2-y1)squared + (z2-z1)squared
Vector equation = (x, y, z) = (xo, yo, zo) + t(a, b, c) *a, b, and c is your vector
EXAMPLE 1: A sphere has points (6, 2, 3) and (2, 2, 7). Find the center and radius.
A). Use the midpoint formula first.
(6+2/2, 2+2/2, 3+7/2) = (4, 2, 5)
B). Now use the IABI formula.
Square root (2-6)squared + (2-2)squared + (7-3)squared = 0
C). The answer you get there is your diameter.
D. Now plug the answer you got from step A into the vector equation formula.
(x-4)squared + (y-2)squared + (z-5)squared = 0 (radius squared)
EXAMPE 2: Find the vector and parametric equations of the line containing (1, 3, 5) and (2, 8, 9)
A). Use your vector equation formula.
(x, y, z) = (1, 3, 5) + t(1, 5, 4)
*(1, 5, 4) is found by subtracting x2-x1, y2-y1, and z2-z1
B). x = 1 + t, y = 3 + 5t, and z = 5 + 4t
Alrighttttttttttttttt, hope everyone has a great weekend and good luck to everyone at Carnival Ball :)
The formula to find midpoint is (x1+x2/2, y1+y2/2, z1+z2/2)
IABI = square root (x2-x1)squared +(y2-y1)squared + (z2-z1)squared
Vector equation = (x, y, z) = (xo, yo, zo) + t(a, b, c) *a, b, and c is your vector
EXAMPLE 1: A sphere has points (6, 2, 3) and (2, 2, 7). Find the center and radius.
A). Use the midpoint formula first.
(6+2/2, 2+2/2, 3+7/2) = (4, 2, 5)
B). Now use the IABI formula.
Square root (2-6)squared + (2-2)squared + (7-3)squared = 0
C). The answer you get there is your diameter.
D. Now plug the answer you got from step A into the vector equation formula.
(x-4)squared + (y-2)squared + (z-5)squared = 0 (radius squared)
EXAMPE 2: Find the vector and parametric equations of the line containing (1, 3, 5) and (2, 8, 9)
A). Use your vector equation formula.
(x, y, z) = (1, 3, 5) + t(1, 5, 4)
*(1, 5, 4) is found by subtracting x2-x1, y2-y1, and z2-z1
B). x = 1 + t, y = 3 + 5t, and z = 5 + 4t
Alrighttttttttttttttt, hope everyone has a great weekend and good luck to everyone at Carnival Ball :)
Sunday, February 13, 2011
Inverse for a 3x3
This week in advanced math we are working on chapter 12. Mrs. Robinson currently showed us how to find the inverse of a 3x3 matrix. This was fairly easy for me because I was already taught it in the number theory class that I am taking, because of this Mrs. Robisnon let me get a head start on my homework (HA HA HA). Anyways I am going to teach you how to do this.
First, you are going to have to pick a row with the lowest numbers in them, then delete the row in column that they are in leaving you with a 2x2 finding the determine of that then multiplying the outside number by the determinate roatating signs postive and negative, starting with positive, and then add all the determinates up.
EXAMPLE:
1. [1 2 2
0 0 1
1 3 0]
0[2 2 0[1 2 1[0 0
3 0] - 1 0]+ 1 3]
0-0+0= 0
so your answer for this one happens to be 0.
First, you are going to have to pick a row with the lowest numbers in them, then delete the row in column that they are in leaving you with a 2x2 finding the determine of that then multiplying the outside number by the determinate roatating signs postive and negative, starting with positive, and then add all the determinates up.
EXAMPLE:
1. [1 2 2
0 0 1
1 3 0]
0[2 2 0[1 2 1[0 0
3 0] - 1 0]+ 1 3]
0-0+0= 0
so your answer for this one happens to be 0.
3-D
This week we learn a lot of matrices and vectors. But this section is 3-D aka the hardest section.
|AB| = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2
Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)
Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2
Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)
Vector addition, magnitude, dot product, etc. all follow the same formula with extra variable.
Example: point 8,-2, 3 and 4, 0, 7
Find center = (6,-1,5)
Radius square root (4-8)^2+(0+2)^2(7-3)^2=16+4+16 = square root of 36= 6
Equation = (x-6)^2+(y+1)^2+(2-5)^2=9^2
Basically if you know your formulas you will do pretty good at this section if not you will do horrible at this section.
|AB| = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2
Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)
Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2
Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)
Vector addition, magnitude, dot product, etc. all follow the same formula with extra variable.
Example: point 8,-2, 3 and 4, 0, 7
Find center = (6,-1,5)
Radius square root (4-8)^2+(0+2)^2(7-3)^2=16+4+16 = square root of 36= 6
Equation = (x-6)^2+(y+1)^2+(2-5)^2=9^2
Basically if you know your formulas you will do pretty good at this section if not you will do horrible at this section.
This week in advanced math we started on chapter 12. We also delved deeper into the Aleks program. I absolutely hate that program because Im the type of person with little time, I dont have all day to sit at a computer because Im never home. In other words I miss blogs so bring them back and get rid of Aleks. But anyway back to chapter 12.
Vectors and Planes
-ax + by + cz+ d when ax + by + cz= d
|a,b,c| must be a vector perpendicular to the plane
3D graphing
-Perpendicular vector to a plane are said to be normal.
Ex. idk if this is right
Vector( 6, 5, -4) is perpendicular to a plane that contains A(0,1,2) Find an equation of the plane.
4x-5y+6x=24
This week we focused more on lessons in class and less on alex. We are finally get back into trig and preparing for our trig exam. Next week our blue pie for alex will be due so we can get into logs and more things with trig. In the section below I will explain our notes on 3D. Chapter 12 has been a lot of formulas and has been pretty simple if you study.
Formulas you need to know for this section are below:
AB = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2
Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)
Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2
Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)
Vector addition, magnitude, dot product, etc.are just a little different but follow the same formulas as earlier sections. They just have an extra variable in the formula unlike before.
Example
Find the center and radius Point A (8,4,2) Point B (2, 8,6)
Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)
Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 6+ 6 + 6= square root 16/2 = square root 8
Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 8
Formulas you need to know for this section are below:
AB = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2
Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)
Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2
Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)
Vector addition, magnitude, dot product, etc.are just a little different but follow the same formulas as earlier sections. They just have an extra variable in the formula unlike before.
Example
Find the center and radius Point A (8,4,2) Point B (2, 8,6)
Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)
Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 6+ 6 + 6= square root 16/2 = square root 8
Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 8
Math 3 dimensions
School is getting old B-Rob. We need to change school to be like 3 days a week, but thanks for not giving us bookoo homework. This week we did some stuff with vectors, except these had a third dimension. Everything we do for them is exactly the same, except you just add an extra set of numbers.
The formula used is- square root (x2-x1)^2+(y2-y1)^2+(z2-z1)^2
Ex: Find |AB| for the points ( 2,3,2) (6, 7, 4)
Square root (6-2) ^2 + (7 – 3) ^2 + (4 – 2)^2
16 + 16 +4 = 36
Square root 36 = 4; |AB| = 6
The formula used is- square root (x2-x1)^2+(y2-y1)^2+(z2-z1)^2
Ex: Find |AB| for the points ( 2,3,2) (6, 7, 4)
Square root (6-2) ^2 + (7 – 3) ^2 + (4 – 2)^2
16 + 16 +4 = 36
Square root 36 = 4; |AB| = 6
3D
This week we did a lot of different things with vectors and matrices. They all seemed pretty hard when they were taught but they are actually not that bad and for the first time I was able to fly through one of the homeworks. Here are the 3D notes, which was probably the hardest section for me:
For magnitude: |AB| = squareroot (x2 – x1)^2 + (y2 – y1)^2 + (z2 + z1)^2
Ex: Find |AB| for the points ( 2, 3), (3, 7), (1, 0)
Square root (3-2) ^2 + (7 – 3) ^2 + (0 – 1)^2
1 + 16 -1 = 16
Square root 16 = 4; |AB| = 4
Ex2: Find |AB| for the points (3, 6) , (3, 0 ) , ( 4, 7)
Square root ( 6 – 3 ) ^2 + ( 0 - 3 ) ^2 + ( 7 – 4 ) ^2
9 -9 + 9 = 9
Square root 9 = 3; |AB| = 3
For magnitude: |AB| = squareroot (x2 – x1)^2 + (y2 – y1)^2 + (z2 + z1)^2
Ex: Find |AB| for the points ( 2, 3), (3, 7), (1, 0)
Square root (3-2) ^2 + (7 – 3) ^2 + (0 – 1)^2
1 + 16 -1 = 16
Square root 16 = 4; |AB| = 4
Ex2: Find |AB| for the points (3, 6) , (3, 0 ) , ( 4, 7)
Square root ( 6 – 3 ) ^2 + ( 0 - 3 ) ^2 + ( 7 – 4 ) ^2
9 -9 + 9 = 9
Square root 9 = 3; |AB| = 3
Ok so this week has been kind of a blur. For some reason I can’t remember exactly what we did this week except I know we learned a couple of new sections. One of these sections has something about 3D. I can’t even remember the section number but anyway it has something to do with vectors and midpoint equations. Here are the notes Mrs. Robinson gave us.
|AB| = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2
Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)
Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2
Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)
Vector addition, magnitude, dot product, etc. all follow the same formula with extra variable.
Example
Find the center and radius Point A (9,5,7) Point B (7, 3,5)
Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)
Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 4 + 4 + 4 = square root 12/2 = square root 3
Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 3
|AB| = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2
Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)
Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2
Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)
Vector addition, magnitude, dot product, etc. all follow the same formula with extra variable.
Example
Find the center and radius Point A (9,5,7) Point B (7, 3,5)
Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)
Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 4 + 4 + 4 = square root 12/2 = square root 3
Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 3
Friday, February 11, 2011
12-1
12-1
Section 12-1 was about vectors. Vectors represent motion and have a magnitude and direction. To add vectors, you add head to tail. To add vectors algebraically, you add components. To find the magnitude of a vector, you use the formula square root of x^2 + y^2. When given two points, such as A and B, and asked to find the vector, you use the formula (x2 – x1, y2 – y1).
Having these few little notes, let’s try an example.
EXAMPLE 1: Given A(6,4) and B(8, 9), find the vector AB.
A). According to the notes above, we’re going to use the last formula.
B). First, subtract the x’s (8-6) = 2
C). Now, subtract the y’s (9-4) = 5
D). Your vector is (2,5)
EXAMPLE 2: Find the magnitude of EXAMPLE 1’S answer. **MAGNITUDE HAS TO BE WITH A VECTOR!
A). Since we’re finding the magnitude, use the square root formula.
B). square root 2^2+5^2 = square root 4 + 25
C). Your magnitude is square root 29
Section 12-1 was about vectors. Vectors represent motion and have a magnitude and direction. To add vectors, you add head to tail. To add vectors algebraically, you add components. To find the magnitude of a vector, you use the formula square root of x^2 + y^2. When given two points, such as A and B, and asked to find the vector, you use the formula (x2 – x1, y2 – y1).
Having these few little notes, let’s try an example.
EXAMPLE 1: Given A(6,4) and B(8, 9), find the vector AB.
A). According to the notes above, we’re going to use the last formula.
B). First, subtract the x’s (8-6) = 2
C). Now, subtract the y’s (9-4) = 5
D). Your vector is (2,5)
EXAMPLE 2: Find the magnitude of EXAMPLE 1’S answer. **MAGNITUDE HAS TO BE WITH A VECTOR!
A). Since we’re finding the magnitude, use the square root formula.
B). square root 2^2+5^2 = square root 4 + 25
C). Your magnitude is square root 29
Sunday, February 6, 2011
Advanced Math
This week in advanced math we learned matrices and we also started to learn chapter 12. Chapter 12 has to do with vectors.
- Vectors represent motion
- Vectors have a magnitude and a direction
- To add vectors you add head to tail.
- The triangle formed by the vectors give the resultant vector
- To add vectors algebraically you add components
To find the magnitude of a vector you do (x2-x1, y2-y1)
Ex. A(11,6) B(12,7) Find vertices AB
(7-6),(12-11)
AB=1,1
Chapter 12
This following week we got back into trig. It was weird and kinda hard to switch back after having such easy lessons the past weeks before. We were introduced to Chapter 12 and also did a lot of work on aleks. It was a hectic week with all the "snow" threats and the rainy weather. I understand Chapter 12-Section 1 more than anything else because there wasn't much to it. It was mostly just having to connect the vectors, add on to them, or flip them around.
vector- is a line that represents motion
vector- is a line that represents motion
- line has a magnitude and a direction, which are the points that are given.
- ( x is the head and y is the tail )
- when is given, it is not asking for an absolute value like usual. that symbol means find the magnitude.
When working with a vector, it usually ask for 3 things.
Example:
Find 4t
- That is telling you to take the vector that it gives you and draw it four times connecting in a straight line.
- If it asks you to find –t, you draw the vector facing the opposite direction.
- Lastly, if it wants you to find t + q, you connect the two vectors together, usually creating an angle.
This past week we did a lot of stuff with aleks. It is really helping me alot. If you do not understand something all you have to do is click explain and it explains it to you. It has helped me a lot with like, simple math and stuff. The only thing i do not like is that it kind of like stalks you because it can tell if you click on another tab and stuff. And it is soooo many problems. But other than that, it is pretty helpful. This week we also learned about vectors. They represent motion. If you are adding vectors you add them from head to tail, but if you are adding them algebraically, you add the components. The triangle that you get after you add the vectors from head to tail is the result of what you added. They have a magnitude and a direction. I can not relaly do any examples because its all drawing graphs and such.
..... Vectors?
All of my days are combining together so, I'm not entirely sure if this is from this week or the week I missed. But here we gooo....
Vectors represent motion. They have a magnitude and direction. They are written with a --> over the two given points.
To add vectors, you must add head to tail. The triangle formed by the vectors give the resultant vector. This basically means that the imaginary line you get when you have
^ \ is the answer.
/ v
To add vectors algebraically, you have to add the components. To do that use this formula: (x (subscript 2)-x (subscript 1), y(subscript 2) - y(subscript 1))
-->
Example-- Find AB A(4,2) B(9,-1)
(9-4, -1-2)
=(5,-3)
To find the magnitude, plug the answer you get from the algebraic vector into this formula: square root(x^2 + y^2). Magnitude is shown as | |
-->
Example-- Find the Magnitude of AB
Square root(5^2 + -3^2)
Square root(25+9)
=Square root(34)
Vectors represent motion. They have a magnitude and direction. They are written with a --> over the two given points.
To add vectors, you must add head to tail. The triangle formed by the vectors give the resultant vector. This basically means that the imaginary line you get when you have
^ \ is the answer.
/ v
To add vectors algebraically, you have to add the components. To do that use this formula: (x (subscript 2)-x (subscript 1), y(subscript 2) - y(subscript 1))
-->
Example-- Find AB A(4,2) B(9,-1)
(9-4, -1-2)
=(5,-3)
To find the magnitude, plug the answer you get from the algebraic vector into this formula: square root(x^2 + y^2). Magnitude is shown as | |
-->
Example-- Find the Magnitude of AB
Square root(5^2 + -3^2)
Square root(25+9)
=Square root(34)
chapter 12
This week was really frustrating because we are getting back into trig after having some pretty easy lessons lately. We covered sections 12-1 and 12-2.
12-1 was pretty easy because all there really was for us to do was connect the vectors, add on to them, or flip them around. A vector is a line that represents motion. The line has a magnitude and a direction, which are the points that are given. ( x is the head and y is the tail ) also, when | | is given, it is not asking for an absolute value like usual. | | that symbol means find the magnitude.
As for the vectors, there are 3 things that it usually asks:
For example: Find 4t
That is telling you to take the vector that it gives you and draw it four times connecting in a straight line.
If it asks you to find –t, you draw the vector facing the opposite direction.
Lastly, if it wants you to find t + q, you connect the two vectors together, usually creating an angle.
I think we should continue to go to the library sometimes to do aleks. I find its really helpful and not as boring as doing work out of the book. Sorry this was such a short blog.
12-1 was pretty easy because all there really was for us to do was connect the vectors, add on to them, or flip them around. A vector is a line that represents motion. The line has a magnitude and a direction, which are the points that are given. ( x is the head and y is the tail ) also, when | | is given, it is not asking for an absolute value like usual. | | that symbol means find the magnitude.
As for the vectors, there are 3 things that it usually asks:
For example: Find 4t
That is telling you to take the vector that it gives you and draw it four times connecting in a straight line.
If it asks you to find –t, you draw the vector facing the opposite direction.
Lastly, if it wants you to find t + q, you connect the two vectors together, usually creating an angle.
I think we should continue to go to the library sometimes to do aleks. I find its really helpful and not as boring as doing work out of the book. Sorry this was such a short blog.
Math
This week we did Aleks a lot. Aleks is great. Every time I get a problem right it tells me how great I am and its a good motivater. We need to spend more CLASS time doing Aleks, not just home time. It wouldn't be so bad for at home but 10 units is a lot B-Rob. I don't really remember what we did to much this week so i'll just put my notes from section 12-1 up hear. BTW we had baseball on Superbowl Sunday which is ridiculous.
vectors represent motion
they have a magnitude and direction.
to add vectors you add head to tail.
the triangle formed by the vectors is your result.
to add vectors algebraically you add the components.
vectors represent motion
they have a magnitude and direction.
to add vectors you add head to tail.
the triangle formed by the vectors is your result.
to add vectors algebraically you add the components.
This week we continued to work on aleks.com. I love aleks because i can work at my own pace and i'm learning a lot. We finished up chapter 14 on matrices (which i absolutely love). Now we are working on vectors. chapter 12-1 and 12-2.
12-1
VECTORS-represent motion
they have a magnitude and direction.
-to add vectors you add head to tail.
-the triangle formed by the vectors is your result.
-to add vectors algebraically you add the components.
12-2
-to find the magnitude
**square root of x^2+y^2
-scalar multiplication puts a scalar times each component.
-you are given 2 points (A&B)
**A&B=(x2-x1;y2-y1)
^^^^remember this from algebra 2!
*it's hard to do examples for 12-1 since they are graphs drawn.
Now im going to spread my love for aleks.com
it is a whole lot easier than doing 2 blogs per week.
im learning from the stupid mistakes i usually do.
its fun?
*i think as a class we all enjoy it. (especially when we do that in class)
byeeeeeeeeeeeeeeee.
12-1
VECTORS-represent motion
they have a magnitude and direction.
-to add vectors you add head to tail.
-the triangle formed by the vectors is your result.
-to add vectors algebraically you add the components.
12-2
-to find the magnitude
**square root of x^2+y^2
-scalar multiplication puts a scalar times each component.
-you are given 2 points (A&B)
**A&B=(x2-x1;y2-y1)
^^^^remember this from algebra 2!
*it's hard to do examples for 12-1 since they are graphs drawn.
Now im going to spread my love for aleks.com
it is a whole lot easier than doing 2 blogs per week.
im learning from the stupid mistakes i usually do.
its fun?
*i think as a class we all enjoy it. (especially when we do that in class)
byeeeeeeeeeeeeeeee.
14-3
This week in advanced math we finished up chapter 14 and took a test and now we are currently working on chapter 12. Chapter 12 is about vectors but I am going to show you how to take the inverse of a matrix today. This is probably one of the easiest things to do when it comes to matricies. Here is how you find an inverse of a matrix.
1. First you are going to want to switch the first number of the matrix with the fourth number.
2. Then make the second and third numbers opposite of their sign.
3. Then cross multiply your origional matrix and find the determinate.
4. Take your determinate and multiply it by your second matrix.
EXAMPLE:
1. [2 3
1 4]
[4 -3
-1 2]
(4x2-1x3)= 5
1/5[4 -3
-1 2]
[4/5 -3/5
-1/5 2/5] This is your answer.
This is the easiest way to do this just follow the steps and you should be fine.
1. First you are going to want to switch the first number of the matrix with the fourth number.
2. Then make the second and third numbers opposite of their sign.
3. Then cross multiply your origional matrix and find the determinate.
4. Take your determinate and multiply it by your second matrix.
EXAMPLE:
1. [2 3
1 4]
[4 -3
-1 2]
(4x2-1x3)= 5
1/5[4 -3
-1 2]
[4/5 -3/5
-1/5 2/5] This is your answer.
This is the easiest way to do this just follow the steps and you should be fine.
12-1,12-2
This week we learned some sections in chapter 12. In 12-1 and 12-2 we learned about vectors and how to represent it.
Vectors-represent motion and it has a magnitude and a direction.
*to add vectors you add head to tail.
the triangle formed by the vectors give the resultant vector.
another thing is to add vector algebraically you add components.
-to find the magnitude
square root of x^2+y^2 magnitude is done as |->|
Ex: A(4,2) B(9,-1) find on top of AB is ->
on top of AB ->(9-4,-1-2)=(5,-3)is on top of AB ->
find the magnitude of on top of AB ->
square root of 5^2+-3^2= square root of 34
Ex: find |on top of AB ->| if A(1,3) B(2,4)
on top of AB -> = 2-1,4-3= (1,1)
|on top of AB ->| Square root of 1^2+1^2= square root of 2.
If you know all your rules and stuff sections 12-1 and 12-2 will be easy for u.
Vectors-represent motion and it has a magnitude and a direction.
*to add vectors you add head to tail.
the triangle formed by the vectors give the resultant vector.
another thing is to add vector algebraically you add components.
-to find the magnitude
square root of x^2+y^2 magnitude is done as |->|
Ex: A(4,2) B(9,-1) find on top of AB is ->
on top of AB ->(9-4,-1-2)=(5,-3)is on top of AB ->
find the magnitude of on top of AB ->
square root of 5^2+-3^2= square root of 34
Ex: find |on top of AB ->| if A(1,3) B(2,4)
on top of AB -> = 2-1,4-3= (1,1)
|on top of AB ->| Square root of 1^2+1^2= square root of 2.
If you know all your rules and stuff sections 12-1 and 12-2 will be easy for u.
Sections 12-1 and 12-2
This week in advanced math we took our test on Wednesday instead of Tuesday because we got out of school early due to the weather. The test was super easy so hopefully I got a good grade. Since we took the test on Wednesday we had to double up on learning sections in chapter 12. Sections 12-1 and 12-2 are on vectors. Vectors represent motion. They have a magnitude and a direction.
To add vectors you add the head and the tail. To add vectors algebracically you add the components.
The triangle formed by the vectors give the resultant vector.
To find the magnitude square root x2 + y2 magnitude is done as | -> |
Scalar multiplication puts a scalar times each component.
Given 2 points A and B
->
AB = (x2 – x1 y2 –y1)
Geometric Representation
Find c + d
Connect the head of C to the tail of D.
Find 2s
Connect the head of S to another tail of S.
Find –p
Reverse the p and draw it in the opposite direction.
This week in advanced math we took our test on Wednesday instead of Tuesday because we got out of school early due to the weather. The test was super easy so hopefully I got a good grade. Since we took the test on Wednesday we had to double up on learning sections in chapter 12. Sections 12-1 and 12-2 are on vectors. Vectors represent motion. They have a magnitude and a direction.
To add vectors you add the head and the tail. To add vectors algebracically you add the components.
The triangle formed by the vectors give the resultant vector.
To find the magnitude square root x2 + y2 magnitude is done as | -> |
Scalar multiplication puts a scalar times each component.
Given 2 points A and B
->
AB = (x2 – x1 y2 –y1)
Geometric Representation
Find c + d
Connect the head of C to the tail of D.
Find 2s
Connect the head of S to another tail of S.
Find –p
Reverse the p and draw it in the opposite direction.
Saturday, February 5, 2011
14-3
Here's my weekend bloggggg:
This week we finished Chapter 14 and tested on it. Along with that we continued to work on Aleks and even started Chapter 12 which had to do with vectors. I’m going to explain the concept of finding an inverse in a matrix.
EXAMPLE 1: Given A= [4 3 find its inverse.
2 6]
A). To find an inverse, flip the 1st number on the top row and the 2nd on the bottom (4 and 6) and make the 2nd number on the first row and 1st on the second row negative (-3 and -2).
B). Your new matrix should look like this [6 -3
-2 4]
C). Now cross multiply your original matrix and subtract (4 x 6 – 2 x 3) = 18
D). Mutiply your new matrix by the determinant, which is 1/18.
E). Your answer is [1/3 -1/6
-1/9 2/9]
I know this is a little short but I've been doing Aleks all morning and I'm really frustrated :/ sorry!
This week we finished Chapter 14 and tested on it. Along with that we continued to work on Aleks and even started Chapter 12 which had to do with vectors. I’m going to explain the concept of finding an inverse in a matrix.
EXAMPLE 1: Given A= [4 3 find its inverse.
2 6]
A). To find an inverse, flip the 1st number on the top row and the 2nd on the bottom (4 and 6) and make the 2nd number on the first row and 1st on the second row negative (-3 and -2).
B). Your new matrix should look like this [6 -3
-2 4]
C). Now cross multiply your original matrix and subtract (4 x 6 – 2 x 3) = 18
D). Mutiply your new matrix by the determinant, which is 1/18.
E). Your answer is [1/3 -1/6
-1/9 2/9]
I know this is a little short but I've been doing Aleks all morning and I'm really frustrated :/ sorry!
Wednesday, February 2, 2011
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