This week we were introduced to Aleks.com which is a website to help better our math skills. I have been working on a lot of old algebra things I have lost my grip on and I like that the website is helping me get back into the hang of the old concepts. So for my blog I am going to go back to Algebra and go over solving a linear equation.
Im just going to do a few examples because im not sure how to explain this concept to well.
Example:
-2= 7x-9/6- 8x+5/7
First we are going to examine the problem and determine the least common demominator.
The least common denominator is 42.
Multiply 42 by each segment of the problem.
(42)-2= 42(7x-9/6-8x+5/7)
(42)-2+ 42(7x-9/6)-42(8x+5/7)
Now divide into the LCD.
-84= 7(7x-9)-6(8x+5)
-84= 49x-63-48x-30
-84=x-93
Now solve for x and
answer is.... x=9
Sunday, January 30, 2011
Matrices are AWESOME
Yeah thats right read the title. It is because of these row by column organized numbers that I will be able to keep my grades up for now. They are simple and easy to do and having something easy to do could not have come at a better time. Baseball is usually a pretty laid-back sport where you go have fun, but since the season has yet to start it has become a boot camp. I am physically and mentally exausted from practicing everday and having a break from hard math is only helping. Matrices are easy to add and subtract. If they are of the same dimensions you simply add or subtract them with their corrresponding number. If not you can't do anything. Multiplication is a little more complicated but not really. The columns of the first matricee has to be equal to the number of rows in the second. Then you take the number of rows in the first and combine it with the number columns with the second and get the dimensions of the product. After all of this you do some fancy math and get the answer. This blog was fun. Not Really. I'm goin to sleep. Night.
Matrices
Well helloo :) my weekend has been going well so far in case you all were wondering. But anyways, this past week we learned about Matrices and we also began the aleks program on the computer. Other than that we pretty much reviewed because we had already learned about matrices in algebra 2. Matrices are pretty simple to work or solve but they are a bit tedious.
There are some rules for matrices:
When adding matrices, they must be of the same dimensions. The same goes for subtracting matrices.
Ex. A 2x3 matrices can only be added to a 2x3 matrices
When multiplying matrices, you can only multiply matrices that have the same colomns and rows.
Ex. 2x3* 3x1 the colomn of the first are equal to the rows of the second.
You cannot technically divide a matrices, to "divide" a matrices, you simply multiply by the inverse..
Ex. inverse
0 1
1 0 = 0-1=-1 = 1/-1* 0 1
1 0 = 0 -1
-1 0
This past week we learned matrices and worked on aleks. Matrices are super easy and stuff.
ADDITION:
[2 6]+[5 4]=[7 10]
SUBTRACTION:
[9 2]-[7 1]=[2 1]
MULTIPLICATION:
5[2 4]=[10 20]
So i'm really tired and I probably shouldn't have waited until the last minute to do my blog but oooops. I guess it will just have to be a short one but next week i'll make it longer :)
ADDITION:
[2 6]+[5 4]=[7 10]
SUBTRACTION:
[9 2]-[7 1]=[2 1]
MULTIPLICATION:
5[2 4]=[10 20]
So i'm really tired and I probably shouldn't have waited until the last minute to do my blog but oooops. I guess it will just have to be a short one but next week i'll make it longer :)
14-1
Section 14-1
This week in advanced math we started chapter 14. Chapter 14 is all about matrices. Matrices are super easy stuff we learned in algebra 2. We finished up our chapter 4 test on Monday and took a quiz either Tuesday or Thursday on matrices but I can’t remember which day. We also started with this new online math website that is kinda like skills tutor. It’s pretty cool I guess, helped me remember some stuff I forgot from algebra 2. Anyway let’s do some super easy problems on matrices.
Examples:
5 [3 6] = [15 30]
[9 10] [45 50]
*this is supposed to be a 2x2 but idk how to do big brackets so all my examples will look pretty
wack so deal with it.
[1 3] + [2 4] = [3 7]
[5 7] [6 8] [11 15]
Name the matrice
[5 6 7]
[4 1 -3]
[3 -1 -4]
3x3
This week in advanced math we started chapter 14. Chapter 14 is all about matrices. Matrices are super easy stuff we learned in algebra 2. We finished up our chapter 4 test on Monday and took a quiz either Tuesday or Thursday on matrices but I can’t remember which day. We also started with this new online math website that is kinda like skills tutor. It’s pretty cool I guess, helped me remember some stuff I forgot from algebra 2. Anyway let’s do some super easy problems on matrices.
Examples:
5 [3 6] = [15 30]
[9 10] [45 50]
*this is supposed to be a 2x2 but idk how to do big brackets so all my examples will look pretty
wack so deal with it.
[1 3] + [2 4] = [3 7]
[5 7] [6 8] [11 15]
Name the matrice
[5 6 7]
[4 1 -3]
[3 -1 -4]
3x3
Matrices
This week we learned how to do matrices. You can add, subtract, multiply, divide, square, cube... ect... matrices. They have other term that we learned to like transpose and trace. But today I am going to teach you how to add, subtract, and transpose matrices.
ADD:
1. You are going to add the first number in the first matrix to the first number in the second matrix.. you do this for the second, third, forth, and so on.. This rule applies the same for subtracting.
Example:
[1 2 [1 2 [2 4
3 4] + 3 4] = 6 8]
This is as easy as it gets.
2. Subtracting- do the same as adding but you subtract!
Example:
[1 2 [1 2 [0 0
3 4]- 3 4] = 0 0]
This is also as easy as it gets.
3. Transpose- when transposing you just make the row a column.
Example:
[1 2 3 4
5 6 7 8
9 9 9 9]
now transpose it.
[1 5 9
2 6 9
3 7 9
4 8 9]
** This are the easiest possible ways to add subtract and transpose matrices.
I hope you all got the hang of it, if not too bad cause you will officially be called an idiot.
ADD:
1. You are going to add the first number in the first matrix to the first number in the second matrix.. you do this for the second, third, forth, and so on.. This rule applies the same for subtracting.
Example:
[1 2 [1 2 [2 4
3 4] + 3 4] = 6 8]
This is as easy as it gets.
2. Subtracting- do the same as adding but you subtract!
Example:
[1 2 [1 2 [0 0
3 4]- 3 4] = 0 0]
This is also as easy as it gets.
3. Transpose- when transposing you just make the row a column.
Example:
[1 2 3 4
5 6 7 8
9 9 9 9]
now transpose it.
[1 5 9
2 6 9
3 7 9
4 8 9]
** This are the easiest possible ways to add subtract and transpose matrices.
I hope you all got the hang of it, if not too bad cause you will officially be called an idiot.
Matrices
This week we learned 14-1 and 14-2. In 14-1 and 14-2 is it all about matrices and how to add, subtract and multiply and determine if they are matrices.
Row and column (row go across and column go up and down)
Examples:
1.
-3{1 4} = {3 12 }
9 6 27 18
2.
-{ 1 1 1] determine row x column= 4x3
1 1 1
0 0 0
2 2 2
3.
A={ 2 6}
8 22 a find a^t
9 16
B. A^t= {2 6}
8 22
9 16
C. { 1 2 }{ 2 4}
4 6 0 1
2x2 2x2
Middle two are the same with you do now
{ 2+0 4+2}
8+0 16+6 ={ 2 6}
8 22
{ 1 2} + {2 0} ={ 3 2}
4 6 5 2 9 8
Basically if u remember everything from last year in algebra 2 it is the same thing and all this is review from last year just know your rule about matrices and you will be fine. Btw my brackets sucked I tried my best making it but oh well hope u like it.
Row and column (row go across and column go up and down)
Examples:
1.
-3{1 4} = {3 12 }
9 6 27 18
2.
-{ 1 1 1] determine row x column= 4x3
1 1 1
0 0 0
2 2 2
3.
A={ 2 6}
8 22 a find a^t
9 16
B. A^t= {2 6}
8 22
9 16
C. { 1 2 }{ 2 4}
4 6 0 1
2x2 2x2
Middle two are the same with you do now
{ 2+0 4+2}
8+0 16+6 ={ 2 6}
8 22
{ 1 2} + {2 0} ={ 3 2}
4 6 5 2 9 8
Basically if u remember everything from last year in algebra 2 it is the same thing and all this is review from last year just know your rule about matrices and you will be fine. Btw my brackets sucked I tried my best making it but oh well hope u like it.
Saturday, January 29, 2011
♥ Marjorie Ann Flanagan St. Martin ♥
This week I did not go to school except for one day, but I didn't stay the whole day and missed 7th hour. But, I did learn lessons this week. I learned how wrong doctors can be. I learned how truly idiotic River Parish Hospital is. I learned how sweet Oschner nurses are. I learned you can survive with one lung and never know the difference. I learned how strong a family can be. I learned how powerful one woman can be. I watched an amazing woman tell a doctor thank you and pat his hand consolingly after he told her she was going to die. I watched this same woman peacefully and gracefully take her last breaths.
Gram was the classiest, sweetest, most wonderful woman to walk this earth. While I miss her dearly and mourn over her loss, I think of how great her life was and how she'll be with her husband again. I think of how happy she must be because there is no doubt about her being in heaven. I love my grandma so much. It's hard to believe she's gone.
Rest In Peace Marjorie Ann Flanagan St. Martin. <3
Gram was the classiest, sweetest, most wonderful woman to walk this earth. While I miss her dearly and mourn over her loss, I think of how great her life was and how she'll be with her husband again. I think of how happy she must be because there is no doubt about her being in heaven. I love my grandma so much. It's hard to believe she's gone.
Rest In Peace Marjorie Ann Flanagan St. Martin. <3
Matrices
Okay so this week we learned matrices and started the aleks stuff on the computer. The first two sections of matrices were easy for me because I remember them from last year. Its going to be super hard to type out matrices because I don’t know how you can put more than one row in a bracket. :/
The basic part is finding out how big your matrices is: rows x colums
(rows go across and columns are up and down)
Ex 1: [ 2 4 6]
This matrix is a 1x3
To add matrices is very simple. You just add the corresponding numbers.
Ex 2: [ 4 2 4] + [ 7 3 6] = [ 11 5 10]
Super easy and subtraction is done the same way.
Ex 3: [ 8 3 2] – [ 5 4 1] = [ 3 -1 1]
Sorry all of my rows were 1, but I have no clue how else to do it!
The basic part is finding out how big your matrices is: rows x colums
(rows go across and columns are up and down)
Ex 1: [ 2 4 6]
This matrix is a 1x3
To add matrices is very simple. You just add the corresponding numbers.
Ex 2: [ 4 2 4] + [ 7 3 6] = [ 11 5 10]
Super easy and subtraction is done the same way.
Ex 3: [ 8 3 2] – [ 5 4 1] = [ 3 -1 1]
Sorry all of my rows were 1, but I have no clue how else to do it!
Matrices!
Woaaaah, feels like I haven’t done one of these in forever!
We started Chapter 14 this week and it’s all about MATRICES! I’m not really sure how I’m going to type them out so bare with me. You can add and subtract matrices but they have to be the same dimensions. Another thing we learned was how to multiply them. When doing so, you have to have the same number of columns in the 1st one and rows in the 2nd one. We also learned how to find its inverse.
Rows – go across
Columns – up and down
Adding and subtracting matrices is simple..
EXAMPLE 1: [-5 0 + [6 -3
4 1] 2 3]
A). To add matrices, we add the corresponding members. Subtraction is done in the same way.
B). (-5+6) (0-3)
(4+2) (1+3)
C). [1 -3
6 4]
D). Your new matrix is a 2 x 2.
EXAMPLE 2: 3[-3 0
4 5]
A). Multiply each number of the matrix by 3, easy.
B). [-9 0
12 15]
When adding, subtracting, and multiplying matrices be careful of negatives! This blog is nasty looking because of those stupid brackets :/
We started Chapter 14 this week and it’s all about MATRICES! I’m not really sure how I’m going to type them out so bare with me. You can add and subtract matrices but they have to be the same dimensions. Another thing we learned was how to multiply them. When doing so, you have to have the same number of columns in the 1st one and rows in the 2nd one. We also learned how to find its inverse.
Rows – go across
Columns – up and down
Adding and subtracting matrices is simple..
EXAMPLE 1: [-5 0 + [6 -3
4 1] 2 3]
A). To add matrices, we add the corresponding members. Subtraction is done in the same way.
B). (-5+6) (0-3)
(4+2) (1+3)
C). [1 -3
6 4]
D). Your new matrix is a 2 x 2.
EXAMPLE 2: 3[-3 0
4 5]
A). Multiply each number of the matrix by 3, easy.
B). [-9 0
12 15]
When adding, subtracting, and multiplying matrices be careful of negatives! This blog is nasty looking because of those stupid brackets :/
Sunday, January 23, 2011
4-5
Last week we learned chapter 4 section 5. in 4-5 we learned about Inverses and how to determine the horizontal line test.
Horizontal line test- similar to the vertical line test only if it passes then there is an inverse that is a function
Sidenote-for our book they will say there isn’t an inverse if it fails
To find an inverse:
-Switch x & y
- solve for y
To check if something is an inverse:
F(g(x))=x
&
g(f(x))=x
-f^-1= inverse notation
Examples:
Prove f(x)=x+3 f^-1(x)=-x-3 are inverse
F(f^-1(X))=x-3+3=x
F^-1(f(X))
(x+3)-3=x
yes it is an inverse.
Y=square root of x-3
x=square root of y-3
square both side you get y-3=x^2
y^-1=x^2+3
it is an inverse because it passes the horizontal line test.
Basically if you know all these formulas and stuff section 4-5 will be easy for you.
Horizontal line test- similar to the vertical line test only if it passes then there is an inverse that is a function
Sidenote-for our book they will say there isn’t an inverse if it fails
To find an inverse:
-Switch x & y
- solve for y
To check if something is an inverse:
F(g(x))=x
&
g(f(x))=x
-f^-1= inverse notation
Examples:
Prove f(x)=x+3 f^-1(x)=-x-3 are inverse
F(f^-1(X))=x-3+3=x
F^-1(f(X))
(x+3)-3=x
yes it is an inverse.
Y=square root of x-3
x=square root of y-3
square both side you get y-3=x^2
y^-1=x^2+3
it is an inverse because it passes the horizontal line test.
Basically if you know all these formulas and stuff section 4-5 will be easy for you.
Inverses Chapter 4
This week we continued Chapter 4. We were introduced to inverse functions in Section 5. This chapter was fairly simple being the only thing you had to do is switch the x’s and y’s in the equation and then simply solve for y. For this section there weren't many steps to the process but before you did anything to the function you must graph the equation on a coordinate plane. Once you have the equation graphed you use a method called the horizontal line test and test the function. If it passes and is an inverse you can move alont the rest of the process.
How to do a horizontal line test: 1.draw horizontal lines across the graph.
2. if it doesn't touch more than one point you can find the inverse.
* Note that in college we will be taught differently and they'll show you a way to find the inverse even if it fails the test.
To show something is an inverse be sure to put the -1 exponent after the y once you have solved the equation.
Example:Find the inverse for y = x -9
You graph this equation, and do the horizontal line test, and it does pass.
1. Switch the x and y. x= y-9
2. Solve for y. y = x + 9
3. Y^-1 = x + 9
How to do a horizontal line test: 1.draw horizontal lines across the graph.
2. if it doesn't touch more than one point you can find the inverse.
* Note that in college we will be taught differently and they'll show you a way to find the inverse even if it fails the test.
To show something is an inverse be sure to put the -1 exponent after the y once you have solved the equation.
Example:Find the inverse for y = x -9
You graph this equation, and do the horizontal line test, and it does pass.
1. Switch the x and y. x= y-9
2. Solve for y. y = x + 9
3. Y^-1 = x + 9
Chapter 4
This week in advanced math we continued in chapter 4. We learned how to decide if a graph was a function and now we learned how to tell if a graph is an inverse. You do this by the vertical line test. We also learned how to find an inverse.
To find and inverse:
1) switch x and y
2) solve for y
If the formula is squared or or taking the absolute value then it is not an inverse.
To check if something is an inverse:
1) do f(g(x))= x and g(f(x))
Also in section 4-4 we learned about stretching and shrinking graphs.
- When you multiply a variable by a number less than 1 it makes the graph shorter and fatter.
- When you multiply a variable by a number bigger than 1 it makes it skinnier and longer.
ALMOST 17 BABY!!!!!!!
That right my b-day is tomorrow. I'm going to be 17 years young. That means i"m old enough to finally get my liscence and drive. How cool is that? Mrs. B-Rob I think you should give me some points as a present. Anywho, we did Math this week. I'm super proud of myself because I believe I did good on most of the tests we were given this week. I can't particularly remember if we learned anything new or not. It's been a pretty long weekend. We did learn some stuff recently about symmetry lately though. This is pretty simple. You have to find if an equations graph is symmetrica on the y-axis, x-axis, the origin, and as well as if y=x.
You may use these steps to perform the process.
x-axis. Step1. Make all y's in the equation negative.
Step2. Simplify the equation. If the equation is the same then it is symmetrical.
y-axis. Step1. Make all x's in the equation negative.
Step2."......"
origin. Step1. Make both the y and x's negative.
Step2. "......"
y=x Step1. Switch the x and ys
Step2. solve for y
Step3. If equations are equal than it is symmetrical.
You may use these steps to perform the process.
x-axis. Step1. Make all y's in the equation negative.
Step2. Simplify the equation. If the equation is the same then it is symmetrical.
y-axis. Step1. Make all x's in the equation negative.
Step2."......"
origin. Step1. Make both the y and x's negative.
Step2. "......"
y=x Step1. Switch the x and ys
Step2. solve for y
Step3. If equations are equal than it is symmetrical.
4-5 Inverses
This week in advanced math we reviewed for a chapter 4 test. We took a took quizzes on Tuesday and Thursday but the quiz on Tuesday didn’t count. Thank you Mrs. Robinson! We took part of the test Friday and we will take the other test on Monday. Section five of chapter four is on inverses. Here is a few notes of inverses:
Horizontal Line Test – similar to the vertical line test only if it passes then there is an inverse that is a function.
To find an inverse – 1. Switch x and y 2. Solve for y
To check if something is an inverse f(g(x)) = x and g(f(x))
f-1 – inverse notation
Side note – In our textbooks it says if there isn’t an inverse if it fails.
Example:
y = x – 7
x = y – 7
y-1 = x + 7
When you draw the graph it passes the horizontal line test.
This week in advanced math we reviewed for a chapter 4 test. We took a took quizzes on Tuesday and Thursday but the quiz on Tuesday didn’t count. Thank you Mrs. Robinson! We took part of the test Friday and we will take the other test on Monday. Section five of chapter four is on inverses. Here is a few notes of inverses:
Horizontal Line Test – similar to the vertical line test only if it passes then there is an inverse that is a function.
To find an inverse – 1. Switch x and y 2. Solve for y
To check if something is an inverse f(g(x)) = x and g(f(x))
f-1 – inverse notation
Side note – In our textbooks it says if there isn’t an inverse if it fails.
Example:
y = x – 7
x = y – 7
y-1 = x + 7
When you draw the graph it passes the horizontal line test.
Chapter 4
In chapter four we learned about functions and how to find their inverses, symmetry, and if they were functions at all.
Finding out if a function is a function at all involves the use of the vertical line test.
When you graph a function; if you can draw a vertical line and have it hit two points at any part of the graph, it's not a function.
The function f(x)=x^2 + 4 would be a function since there's no point on the graph where a vertical line would hit a x point twice.
The graph x=2 is not a function because a vertical line hits all the points on the graph.
Finding a function's inverse
If a function has an inverse, a horizontal line will not be able to touch any two points of the graph.
The graph f(x)= x+3 has an inverse because a horizontal line would cross its graph once.
The graph f(x)=x^2-5 wouldn't have an inverse that is also a function because a horizontal line can touch the graph twice.
Chapter 4
This week in advanced math we learned chapter 4. This was probably one of the easiest chapters that we have done all year. Well Friday we took the multiple choice portion of the chapter test and now Monday we will be taking the free response portion of the test. I think that I did really well on the multiple choice portion of the test. Hopefully the free response test will be just as easy. Now lets talk about INVERSES.
First: You should graph the equation and make sure it passes the horizontal line test.
(this is so you do not waste time finding an inverse of something that isn't mathematically possible.)
Next: If an inverse, you should switch the x and y and solve for y.
**HINTS:
*If it is x^2 it wont have an inverse because it will be a parabola. (no inverse)
*If it is x it will be a straight line (has an inverse)
* If it is x^3 it will make this weird curve line thingy (has an inverse)
EXAMPLE:
1. y=x^2+4
answer: no inverse (this is a parabola)
2. y=x-4
x=y-4
y^1=x+4 is your answer
I hope that I explained this well and if you think I didn't then you are wrong because I did. Thanks for taking up 5 minutes of my time for 5 extra points. :)
First: You should graph the equation and make sure it passes the horizontal line test.
(this is so you do not waste time finding an inverse of something that isn't mathematically possible.)
Next: If an inverse, you should switch the x and y and solve for y.
**HINTS:
*If it is x^2 it wont have an inverse because it will be a parabola. (no inverse)
*If it is x it will be a straight line (has an inverse)
* If it is x^3 it will make this weird curve line thingy (has an inverse)
EXAMPLE:
1. y=x^2+4
answer: no inverse (this is a parabola)
2. y=x-4
x=y-4
y^1=x+4 is your answer
I hope that I explained this well and if you think I didn't then you are wrong because I did. Thanks for taking up 5 minutes of my time for 5 extra points. :)
4-5
This week was weird. I really don't remember too much of what went on in school. Friday, we took a test and the other days were review, I believe. One of the sections on the test was section 4-5. Section 4-5 was about finding inverses.
To find an inverse you have to use the horizontal line test. This is like the vertical line test, but it's horizontal. If it passes without touching two points on the graph then there is an inverse that is a function.
This brings us to the first step - drawing the graph. Since I don't have a fancy laptop, I can't draw a graph. But, here are some hints to see if it will pass the test. If it has an x^2, it will not pass the test because x^2 makes a parabola. An x makes a straight kinda slanted line.
After you graph, to find an inverse switch x and y. Then solve for y.
For example, y = x + 3
First, graph the equation.
By using the horizontal line test, you know that there is an inverse.
Now, switch the x and y.
x = y + 3
And now, solve for y.
y^-1 = x - 3
Place -1 above y to show that it is the inverse.
To find an inverse you have to use the horizontal line test. This is like the vertical line test, but it's horizontal. If it passes without touching two points on the graph then there is an inverse that is a function.
This brings us to the first step - drawing the graph. Since I don't have a fancy laptop, I can't draw a graph. But, here are some hints to see if it will pass the test. If it has an x^2, it will not pass the test because x^2 makes a parabola. An x makes a straight kinda slanted line.
After you graph, to find an inverse switch x and y. Then solve for y.
For example, y = x + 3
First, graph the equation.
By using the horizontal line test, you know that there is an inverse.
Now, switch the x and y.
x = y + 3
And now, solve for y.
y^-1 = x - 3
Place -1 above y to show that it is the inverse.
So this past week we went over a few things from the week before for our chapter test and we learned some new things. Section 4-5 inverses. These are fun and easy :)
INVERSES:
****FIRST THING YOU MUST DO IS GRAPH!
*use the horizontal line test to see if it is an inverse. If it only goes through the line once it is an inverse. if it goes through the line more than once it isnt an inverse.
**switch x and y.
*solve for y.
I left my notes at home so im doing this off the top of my head. I hope this stuff is right. And leaving my notebook at home does not help the fact that we have the other part of the chapter test tomorrow :(
I really do not want to go to school tomorrow. Ive been doing schoolwork allllllllll day and i dont think i will ever finish. It makes me really sad that this is what school has done to me. Free time? what is that? All it does is stress me out!
INVERSES:
****FIRST THING YOU MUST DO IS GRAPH!
*use the horizontal line test to see if it is an inverse. If it only goes through the line once it is an inverse. if it goes through the line more than once it isnt an inverse.
**switch x and y.
*solve for y.
I left my notes at home so im doing this off the top of my head. I hope this stuff is right. And leaving my notebook at home does not help the fact that we have the other part of the chapter test tomorrow :(
I really do not want to go to school tomorrow. Ive been doing schoolwork allllllllll day and i dont think i will ever finish. It makes me really sad that this is what school has done to me. Free time? what is that? All it does is stress me out!
4-5
We learned section 4-5 to complete the rest of chapter 4 before we go back to trig. It was all about finding inverses of a problem. It was very simple and easy because all you have to do is switch the x’s and y’s in a problem and then solve for y. There are very few steps and it is pretty simple. BEFORE you solve the problem, you need to graph the equation on a coordinate plane and use the horizontal line test.
The horizontal line test is where you draw horizontal lines across a graph. If it does NOT touch more than one point, then it is an inverse. According to the book, if it does not pass this test, then an inverse does not exist. We did learn though that in college, we will be told something different. Also, when you find the inverse, you put an exponent of -1 after the y to show that it is an inverse.
Once you passed the horizontal line test, you can get to working the problem:
Example:
Find the inverse for y = x -9
You graph this equation, and do the horizontal line test, and it does pass.
1. Switch the x and y. x= y-9
2. Solve for y. y = x + 9
3. Y^-1 = x + 9
The horizontal line test is where you draw horizontal lines across a graph. If it does NOT touch more than one point, then it is an inverse. According to the book, if it does not pass this test, then an inverse does not exist. We did learn though that in college, we will be told something different. Also, when you find the inverse, you put an exponent of -1 after the y to show that it is an inverse.
Once you passed the horizontal line test, you can get to working the problem:
Example:
Find the inverse for y = x -9
You graph this equation, and do the horizontal line test, and it does pass.
1. Switch the x and y. x= y-9
2. Solve for y. y = x + 9
3. Y^-1 = x + 9
4-5
well to be honest i really dont remember much of this week. This is probally due to the four day weekend we had. We took a test friday, but i really dont remember anything before that. But anywho, i do remember section four five,. This was one if the harder sections of the chapter. Section four five is on inverses.The rules that you have to follow are simple..
1.Draw or sketch the graph
- x^2= parabola
-x= a line
-x^3 =
this line..
-sqrt x=
that thing..
- ab. value = "v"
* if the graph passes the horoizontal line test, it has an inverse.
2. next you switch the x's and the y's
3. Solve for y
ex.
f(x)=x^2+3x
1.has no inverse because it does not pass the horizontal line test.
ex. y=5x+3
1. passes the horizontal line test
2.x=5y+3
x-3=5y
y(-1)=1/5x-3/5 <---inverse
ex. y=l3xl -6
1. does not pass horizontal line test
1.Draw or sketch the graph
- x^2= parabola
-x= a line
-x^3 =
this line..-sqrt x=
that thing..- ab. value = "v"
* if the graph passes the horoizontal line test, it has an inverse.
2. next you switch the x's and the y's
3. Solve for y
ex.
f(x)=x^2+3x
1.has no inverse because it does not pass the horizontal line test.
ex. y=5x+3
1. passes the horizontal line test
2.x=5y+3
x-3=5y
y(-1)=1/5x-3/5 <---inverse
ex. y=l3xl -6
1. does not pass horizontal line test
Friday, January 21, 2011
4-5
Inverses
Section 4-5 was about finding inverses. To find an inverse you have to use the horizontal line test. This is similar to the vertical line test. If it passes without touching two points then there is an inverse that is a function. That is the first step - drawing the graph.
After that, to find an inverse:
1. Switch x and y.
2. Solve for y.
To check if something is an inverse:
f(g(x)) = x and g(f(x)) = x
EXAMPLE 1: y = x + 4
A). First, graph the equation.
B). By using the horizontal line test, we know that there is an inverse.
C). Now, switch the x and y.
D). x = y + 4 and solve for y.
E). y^-1 = x - 4
* Place -1 above y to show that it is the inverse.
EXAMPLE 2: Prove that f(x) = x - 2 and f^-1(x) = x + 2 are inverses.
A). f(f^-1(x)) = (x + 2) - 2
= x
B). f^-1(f(x)) = ( x - 2) + 2
= x
C). This proves that these are inverses.
Tuesday, January 18, 2011
Week 3 Blog Prompt
When you are finding the domain and range of a problem what is it telling you about the graph?
Monday, January 17, 2011
Chapter 4
This week in advanced math we were learning chapter 4 which isn't that hard. But you do need to know the steps and the different types of problems to use in order to succeed. You also need to know things like the line test to tell if a graph is a function or not. And you need to know how to find zeroes.
Horizontal line test
* The horizontal line test is used to tell if a graph is a function.
If you make a straight line going vertical and the vertical line touches the graphed line twice then it is not a funcion. In other words if X is repeated then it is not a function.
Ex: (3,3) (4,3) (5,3) and (6,3) are the points then this is a function because X doesn't repeat itself.
On the other hand (3,3) (4,3) (5,3) and (3,6) would not be a function because x repeats itself therefore it would not pass the vertical line test.
Sunday, January 16, 2011
blogg..
Well, I have not been feeling too well this weekend and i almost forgot about blogging! so that is why i am so late, but anyways.. This week Mrs. Robinson was out a lot , I think it was two days. While she was out we did worksheets on domain and range; which is not so bad, and on factoring which is pretty simple. Ill blog on Domain and Range because many people seemed to have problems with this.
Ex.
x^5+3x-2
first, we figure out whether the equation is a polynomial, an absolute value, a fraction, or a square root. Because the equation does not have a variable in the denominator it therefore is a polynomial.
the domain of a polynomial is always (-infinity, infinity) <-- you have to pay attention to the parenthesis also. Parenthesis only go with infinities and fractions.
The Domain of example one is (-infinity, infinity)
the range is also (-infinity, infinity)
when the exponent of the polynomial ( the greater one) is even the range is (-infinity, infinity) it it is even, then it would be the [number, then infinity)
ex.2 l-4xl +3
the domain of an absolute value is (-infinity, infinity)
the range is the [vertical shift, infinity)
Ex.
x^5+3x-2
first, we figure out whether the equation is a polynomial, an absolute value, a fraction, or a square root. Because the equation does not have a variable in the denominator it therefore is a polynomial.
the domain of a polynomial is always (-infinity, infinity) <-- you have to pay attention to the parenthesis also. Parenthesis only go with infinities and fractions.
The Domain of example one is (-infinity, infinity)
the range is also (-infinity, infinity)
when the exponent of the polynomial ( the greater one) is even the range is (-infinity, infinity) it it is even, then it would be the [number, then infinity)
ex.2 l-4xl +3
the domain of an absolute value is (-infinity, infinity)
the range is the [vertical shift, infinity)
4-3
This past week in advanced math Mrs. Robinson was out for two days so we were given some worksheets to work on. Looking back to the week before last we learned some more of chapter four. Let’s review section 4-3 which deals with symmetry. There are a few rule you need to know:
Check to see if:
X - axis – 1) Plug in (-y)
2) Simplify and solve for y
3) If equations are equal then it has symmetry about x.
Y – axis – 1) Plug in (-x)
2) “ “
3) “ “ y – axis
Origin – 1) Plug in (-x) and (-y)
2) “ “
3) “ “ origin
Y = X – 1) Switch x and y
2) Solve for y
3) “ “ y = x
To graph these types of equations you will do 1 of 3 things:
A) Sketch – f(x) Reflect the x-axis
B) Sketch – f(-x) Reflect the y-axis
C) Sketch - |f(x)| Flip above the axis
Check to see if:
X - axis – 1) Plug in (-y)
2) Simplify and solve for y
3) If equations are equal then it has symmetry about x.
Y – axis – 1) Plug in (-x)
2) “ “
3) “ “ y – axis
Origin – 1) Plug in (-x) and (-y)
2) “ “
3) “ “ origin
Y = X – 1) Switch x and y
2) Solve for y
3) “ “ y = x
To graph these types of equations you will do 1 of 3 things:
A) Sketch – f(x) Reflect the x-axis
B) Sketch – f(-x) Reflect the y-axis
C) Sketch - |f(x)| Flip above the axis
Tired
Yeah I'm not in the mood to do a blog, but my parents say if they keep seeing 0's on edline, well i can't tell you because that's how horrible it will be. Anyway, back to Math. This week Mrs. B-Rob wasn't here for most of the week yet she still managed to give us a quiz :). Fortunately, this quiz was generally kind of easy except for the domain and Range part. I kind of forgot to pay attention that day. We also learned some other stuff about symmetry. This is pretty easy as well, I remembered to pay attention that day, but like all Math the problems can be made extremely difficult by adding in other stuff like extra letters and exponents, etc. Hopefully this week is a week I feel like paying attention so that I can do good on tests. Susan that sucks about your computer, Dance Team I hope ya'll did good, and everyone else, PEACE.
4-3
This week in school we learned chapter 4 section 3. In 4-3 we are learned about symmetry and how to plug it in on a graph. I can have example in this sections cause there is no way for me to make a graph and stuff. I will explain how to do the problems in section 4-3.
x axis: plug in (-y), simplify( solve for y), if the equation are = then it has symmetry a/b x.
If it gives you a graph and tells you to sketch -f(x), you take the graph and draw everything on the x axis on the opposite side.
y axis: plug in (-x),simplify (solve for x), If equations are equal then it has symmetry on the y axis.
If it tells you to sketch f(-x) then you take everything on one side of the y axis and draw it to the other.
|f(x)|, flip above axis
1. plug in (-x) & (-y)
2. simplify
If it tells you to sketch |f(x)| then everything on the lower half of the axis, you draw onto the top half.
y=x is also a inverse.
Ms.Robinson this is all i can explain lmao this section is soo confusing that is why i did it to understand it more.
x axis: plug in (-y), simplify( solve for y), if the equation are = then it has symmetry a/b x.
If it gives you a graph and tells you to sketch -f(x), you take the graph and draw everything on the x axis on the opposite side.
y axis: plug in (-x),simplify (solve for x), If equations are equal then it has symmetry on the y axis.
If it tells you to sketch f(-x) then you take everything on one side of the y axis and draw it to the other.
|f(x)|, flip above axis
1. plug in (-x) & (-y)
2. simplify
If it tells you to sketch |f(x)| then everything on the lower half of the axis, you draw onto the top half.
y=x is also a inverse.
Ms.Robinson this is all i can explain lmao this section is soo confusing that is why i did it to understand it more.
Blah
So, here I am, finally remembering to do my blog. Sigh. I really don't understand why my memory is so terrible.
Anyway, this week was fairly simple in Advanced Math. This section had to do with notation. There are four formulas to find notation: sum, difference, product, and quotient of functions.
1. Sum of f and g: (f+g)(x) = f(x) + g(x)
2. Difference of f and g: (f-g)(x) = f(x) – g(x)
3. Product of f and g: (f * g) g)(x) = f(x) * g(x)
4. Quotient of f and g: (f/g)(x) = f(x)/g(x), provided g(x) not equaled to 0
All this is basically saying is that in the problems that follow, plug everything in the parenthesis of f or g into the equation where there is a variable
For example,
f(3)= 7x= 21
You'd then plug 3 in where there's a x.
7 * 3 = 21
Ta da
For another example
g(4)= 5-x= 9
Then you plug in 4 for the x.
5 - 4 = 9
Anyway, this week was fairly simple in Advanced Math. This section had to do with notation. There are four formulas to find notation: sum, difference, product, and quotient of functions.
1. Sum of f and g: (f+g)(x) = f(x) + g(x)
2. Difference of f and g: (f-g)(x) = f(x) – g(x)
3. Product of f and g: (f * g) g)(x) = f(x) * g(x)
4. Quotient of f and g: (f/g)(x) = f(x)/g(x), provided g(x) not equaled to 0
All this is basically saying is that in the problems that follow, plug everything in the parenthesis of f or g into the equation where there is a variable
For example,
f(3)= 7x= 21
You'd then plug 3 in where there's a x.
7 * 3 = 21
Ta da
For another example
g(4)= 5-x= 9
Then you plug in 4 for the x.
5 - 4 = 9
4-3
sooo my computer has a virus and wont let mee open up anything other than the internet, so I have no idea how many words this is going to be.
Pretty much everything except for 4-2 that we have learned this week has to do with drawing graphs, and since I already did a blog on that and I obviously can't draw any graphs on here, I'll just explain how to work some problems from section 4-3. This section was all about symmetry. Some problems gave you a graph and you had to either sketch:
-f(x), reflect the x axis
1. plug in (-y)
2. simplify (solve for y)
3. if eqations are equal then it has symmetry about x.
* If it gives you a graph and tells you to sketch -f(x), you take the graph and draw everything on the x axis on the opposite side.
f(-x), reflect the y axis
1. plug in (-x)
2. simplify (solve for x)
3. If equations are equal then it has ymmetry on the y axis.
* If it tells you to sketch f(-x) then you take everything on one side of the y axis and draw it to the other.
|f(x)|, flip above axis
1. plug in (-x) & (-y)
2. simplify
* If it tells you to sketch |f(x)| then everything on the lower half of the axis, you draw onto the top half.
Sorry this was confusing, but it is hard to explain it without having some graphs as examples.
Pretty much everything except for 4-2 that we have learned this week has to do with drawing graphs, and since I already did a blog on that and I obviously can't draw any graphs on here, I'll just explain how to work some problems from section 4-3. This section was all about symmetry. Some problems gave you a graph and you had to either sketch:
-f(x), reflect the x axis
1. plug in (-y)
2. simplify (solve for y)
3. if eqations are equal then it has symmetry about x.
* If it gives you a graph and tells you to sketch -f(x), you take the graph and draw everything on the x axis on the opposite side.
f(-x), reflect the y axis
1. plug in (-x)
2. simplify (solve for x)
3. If equations are equal then it has ymmetry on the y axis.
* If it tells you to sketch f(-x) then you take everything on one side of the y axis and draw it to the other.
|f(x)|, flip above axis
1. plug in (-x) & (-y)
2. simplify
* If it tells you to sketch |f(x)| then everything on the lower half of the axis, you draw onto the top half.
Sorry this was confusing, but it is hard to explain it without having some graphs as examples.
Friday, January 14, 2011
So i'd rather not spend my friday night blogging but i wont be home all weekend so here i go.....
4-2 was notation:
1. sum of f and g=(f+g)(x)=f(x)+g(x)
2. difference of f and g=(f-g)(x)=f(x)-g(x)
3. product of f and g=(f x g)(x)=f(x)-g(x)
4. quotient of f and g=(f/g)(x)=f(x)/g(x)
most fuctions of notation are composition functions: which is a function inside another function.
4-3 symmetry:
*you get to draw graphs!
x-axis:
1. plug in (-y)
2. simplify=solve for y
3. if equations are equal then it has symmetry about x.
y-axis:
1. plug in (-x)
2. simplify=solve for y
3. if equations are equal then it has symmetry about y.
origin:
1. plug in (-x) and (-y)
2. simplify=solve for y
3. if equations are equal then it has symmetry of the origin.
y=x
1. switch x and y
2. solve for y
3. if equations are equal then it has symmetry for y=x
*when you draw these graphs its all about reflecting.
EVERYONE ENJOY YOUR 3 DAY WEEKEND :)
4-2 was notation:
1. sum of f and g=(f+g)(x)=f(x)+g(x)
2. difference of f and g=(f-g)(x)=f(x)-g(x)
3. product of f and g=(f x g)(x)=f(x)-g(x)
4. quotient of f and g=(f/g)(x)=f(x)/g(x)
most fuctions of notation are composition functions: which is a function inside another function.
4-3 symmetry:
*you get to draw graphs!
x-axis:
1. plug in (-y)
2. simplify=solve for y
3. if equations are equal then it has symmetry about x.
y-axis:
1. plug in (-x)
2. simplify=solve for y
3. if equations are equal then it has symmetry about y.
origin:
1. plug in (-x) and (-y)
2. simplify=solve for y
3. if equations are equal then it has symmetry of the origin.
y=x
1. switch x and y
2. solve for y
3. if equations are equal then it has symmetry for y=x
*when you draw these graphs its all about reflecting.
EVERYONE ENJOY YOUR 3 DAY WEEKEND :)
4-2
I’m posting this super early, but it’s because I won’t be home till late Sunday night! Anyways, Section 4-2 dealt with notation. There are four formulas to find notation: sum, difference, product, and quotient of functions.
1. Sum of f and g: (f+g)(x) = f(x) + g(x)
2. Difference of f and g: (f-g)(x) = f(x) – g(x)
3. Product of f and g: (f x g)(x) = f(x) x g(x)
4. Quotient of f and g: (f/g)(x) = f(x)/g(x), provided g(x) not equaled to 0
*f(x) is a notation!
EXAMPLE 1: f(x) = 2x + 1 and g(x) = 2 – x Find the sum and difference notations.
A). (f+g)(x) = (2x+1) + (2-x)
= x + 3
B). (f-g)(x) = (2x+1) – (2-x)
= 3x -1
EXAMPLE 2: If you have f(#), f(y), or f(i^2), the notations means to plug what is in the parenthesis into the equation instead of x. Knowing this, solve (f+g)(3) using Example 1’s f(x) and g(x).
A). (f+g)(3) = (2x+1) + (2-x)
= x + 3
B). Plug in 3 into the above answer.
= 3 + 3
C). Your answer is 6.
1. Sum of f and g: (f+g)(x) = f(x) + g(x)
2. Difference of f and g: (f-g)(x) = f(x) – g(x)
3. Product of f and g: (f x g)(x) = f(x) x g(x)
4. Quotient of f and g: (f/g)(x) = f(x)/g(x), provided g(x) not equaled to 0
*f(x) is a notation!
EXAMPLE 1: f(x) = 2x + 1 and g(x) = 2 – x Find the sum and difference notations.
A). (f+g)(x) = (2x+1) + (2-x)
= x + 3
B). (f-g)(x) = (2x+1) – (2-x)
= 3x -1
EXAMPLE 2: If you have f(#), f(y), or f(i^2), the notations means to plug what is in the parenthesis into the equation instead of x. Knowing this, solve (f+g)(3) using Example 1’s f(x) and g(x).
A). (f+g)(3) = (2x+1) + (2-x)
= x + 3
B). Plug in 3 into the above answer.
= 3 + 3
C). Your answer is 6.
Sunday, January 9, 2011
4-1
This week in advanced math we started chapter 4, we dealt with section 4-1 which was Domain and Range. We also had the vertical line test. But 4-1 involves 3 different types of problems. Polynomials, Square Roots, and Fractions. The secret behind chapter 4 is that you have to pay attention to see which formula you have to use.
4-1 Polynomial
-Domain is always (-∞, ∞)
-Range os (-∞, ∞) if odd
Ex: f(x)= 25-5x; solve for x and find domain and range
0= 25-5x
5x=25
x=5
D: (-∞, ∞)
R: (-∞, ∞)
4-2
This week we learn Chapter 4 stuff. In section 4-2 we learn operations on function better known as notation. We have been taught this section from algebra 2 last year. You have to solve for F(x) and G(x). It took me awhile to realize it but it pretty much easy.
A composition function is a function inside of another function. A example of this is F(x).
These are the formulas:
Sum- (f+g)(x)=f(x)+g(x)
Difference- (f-g)(x)=f(x)-g(x)
Product- (f*g)(x)=f(x)*g(x)
Quotient- (f/g)(x)=f(x)/g(x)
The notation is f(x) when f(5) or f(x) or f(i^2) the notation means to plug what is in the parenthesis into the equation instead of x.
Example:
F(X)=3x+1 G(x)=2-x
(f+g)(x)(3x+1)+(-2+x)
4x-1 is the answer.
A example of a composition function is (fog)(x)=f(g(x))-> Write g(x) inside of x. And vise verse for g of f of x.
Basically if you know your formulas and everything this section will be a breeze for you.
A composition function is a function inside of another function. A example of this is F(x).
These are the formulas:
Sum- (f+g)(x)=f(x)+g(x)
Difference- (f-g)(x)=f(x)-g(x)
Product- (f*g)(x)=f(x)*g(x)
Quotient- (f/g)(x)=f(x)/g(x)
The notation is f(x) when f(5) or f(x) or f(i^2) the notation means to plug what is in the parenthesis into the equation instead of x.
Example:
F(X)=3x+1 G(x)=2-x
(f+g)(x)(3x+1)+(-2+x)
4x-1 is the answer.
A example of a composition function is (fog)(x)=f(g(x))-> Write g(x) inside of x. And vise verse for g of f of x.
Basically if you know your formulas and everything this section will be a breeze for you.
4-1
Section 4-1
Domain and Range
This week in advanced math we went back to chapter four since we skipped over it in the first semester. Chapter 4 deals with more algebra than advanced math. We only got started on two sections this week so I’m guessing the whole chapter is on domain and range. Let’s review section 4-1 shall we.
A few simple way to find domain and range
1. Polynomial – an equation with no variables in the denominator.
Domain…. Is always (-infinity, infinity)
Range….. is (-infinity, infinity) if it is odd
2. Square roots –
Domain….. set inside equal to 0 and solve for x, set up a number line, plug in numbers and use the
Positive intervals
Range……. [vertical shift, infinity)
3. Fractions –
Domain……set bottom equal to 0 and solve for x, set up intervals
Range…..take the limit as x -> infinity, set up intervals
Example:
6x5 + 7x is a polynomial so the answer is D: (-infinity, infinity) R: (-infinity, infinity)
Domain and Range
This week in advanced math we went back to chapter four since we skipped over it in the first semester. Chapter 4 deals with more algebra than advanced math. We only got started on two sections this week so I’m guessing the whole chapter is on domain and range. Let’s review section 4-1 shall we.
A few simple way to find domain and range
1. Polynomial – an equation with no variables in the denominator.
Domain…. Is always (-infinity, infinity)
Range….. is (-infinity, infinity) if it is odd
2. Square roots –
Domain….. set inside equal to 0 and solve for x, set up a number line, plug in numbers and use the
Positive intervals
Range……. [vertical shift, infinity)
3. Fractions –
Domain……set bottom equal to 0 and solve for x, set up intervals
Range…..take the limit as x -> infinity, set up intervals
Example:
6x5 + 7x is a polynomial so the answer is D: (-infinity, infinity) R: (-infinity, infinity)
4-2
Im going to refresh our memory on what we learned thursday. We went back into the earlier chapters of the book this week and studied Chapter 4. We were taught to solve for f and g of (x) which we had previously been introduced to last year in Algebra II. It was pretty easy and came back to my memory pretty quick.
Composition- a function inside of another function.
Ex: (f o g) x
Formulas to remember are:
1.Sum- (f+g)(x)=f(x)+g(x)
2.Difference- (f-g)(x)=f(x)-g(x)
3.Product- (f*g)(x)=f(x)*g(x)
4.Quotient- (f/g)(x)=f(x)/g(x)
Plug in the given into the variables place in the equation. This process is pretty basic, being all you really are doing is plugging numbers and solving to simplest form.
Example:
(Difference) <-----Formula
(f-g)(x)=f(x)-g(x)
f(x)=4x-2 g(x)=6x-4
These are your givens, so simply take them and plug them into the right side of the formula.
(f-g)(x)=(4x-2) - (6x-4)
Answer- (-2x+6)
Composition- a function inside of another function.
Ex: (f o g) x
Formulas to remember are:
1.Sum- (f+g)(x)=f(x)+g(x)
2.Difference- (f-g)(x)=f(x)-g(x)
3.Product- (f*g)(x)=f(x)*g(x)
4.Quotient- (f/g)(x)=f(x)/g(x)
Plug in the given into the variables place in the equation. This process is pretty basic, being all you really are doing is plugging numbers and solving to simplest form.
Example:
(Difference) <-----Formula
(f-g)(x)=f(x)-g(x)
f(x)=4x-2 g(x)=6x-4
These are your givens, so simply take them and plug them into the right side of the formula.
(f-g)(x)=(4x-2) - (6x-4)
Answer- (-2x+6)
4-2
This week we went back to chapter 4 in the book which is algebra 2. For 4-2,we learned how to solve for f and g of (x). I pretty much remembered it from last year, but as with anything else, when I go to do it on my own, I get lost. So, like always, there are formulas to remember:
A composition function is a function inside of another function. Such as f(x).
These are the formulas:
Sum- (f+g)(x)=f(x)+g(x)
Difference- (f-g)(x)=f(x)-g(x)
Product- (f*g)(x)=f(x)*g(x)
Quotient- (f/g)(x)=f(x)/g(x)
The variable inside of the parenthesis is where you plug in what you are told into the equation. I think the problems are a bit simpler than the directions make it seem. All you really are doing is plugging in numbers into an equation that is already there and solving.
Ex. (sum)
f(x) = 3x + 5 g(x) = 9x + 8
Since it says to do the sum formula, you find it up at the top and plug in the equations where it tells you to.
(f+ g)(x) = (3x + 5) + (9x + 8)
Simply solve.
12x + 13
A composition function is a function inside of another function. Such as f(x).
These are the formulas:
Sum- (f+g)(x)=f(x)+g(x)
Difference- (f-g)(x)=f(x)-g(x)
Product- (f*g)(x)=f(x)*g(x)
Quotient- (f/g)(x)=f(x)/g(x)
The variable inside of the parenthesis is where you plug in what you are told into the equation. I think the problems are a bit simpler than the directions make it seem. All you really are doing is plugging in numbers into an equation that is already there and solving.
Ex. (sum)
f(x) = 3x + 5 g(x) = 9x + 8
Since it says to do the sum formula, you find it up at the top and plug in the equations where it tells you to.
(f+ g)(x) = (3x + 5) + (9x + 8)
Simply solve.
12x + 13
4-2 Notation
This past week we have taken a semi-break from the trig in our book. We actually went back to chapter four where we began finding the Domain and Range in section one. We also learned in section two about Notation. Section two deals with notation and functions.
Rules:
Sum of f,g (f+g)x= f(x)+g(x)
Differemce of f,g (f-g)= f(x)-g(x)
Product f,g (fxg)=f(x)xg(x)
Quotient f,g (f/g)x= f(x)/g(x)
One of the important kinds of functions that we learned about were the composition functions. A composition function is simply a function inside of another function. A function can be stated as f(x) of as "f of x" with a given equation you simply plug in.
Ex. Difference of a function
F(x)=2x+2 g(x)=4x
(f-g)(x)=f(x)-g(x)= (2x+2)-(4x)= -2x+2
Ex. Composition Function
(f·g)(x)=f(g(x)) this is stated as f of g of x..
= f(4x)
= 2(4x)+2
= 8x+2
That is the final answer. 8x+2
This section is relatively simple. The hardest thing that you will probally have to do is solve. Other than that, all you are doing is plugging in functions.
:)
4-2
This week in advanced math we are learning how to do solve for f and g of (x). I think that this section is very confusing along with section 4-1. Anyways I am going to do my best to explain this to you.
Composition Functions- A composition function is a function inside of another function.
Example: (fog)(x)=f(g(x))-> Write g(x) inside of x. And vise verse for g of f of x.
Sum- (f+g)(x)=f(x)+g(x)
Difference- (f-g)(x)=f(x)-g(x)
Product- (f*g)(x)=f(x)*g(x)
Quotient- (f/g)(x)=f(x)/g(x)
***REMEMBER THIS: f(x) is the notation when you use f(5) or f(y) 0r f(i^2). The notation means to plug what is in parenthesis into the equation instead of x.
EXAMPLE:
f(x)=6x+1 g(x)=3+x
(f+g)(x)=(6x+1)+(3+x)
Answer: 7x+4
This is the easiest way that I can teach you how to do this. It is just plugging into formulas.
Thanks and have a GREAT day. :)
Composition Functions- A composition function is a function inside of another function.
Example: (fog)(x)=f(g(x))-> Write g(x) inside of x. And vise verse for g of f of x.
Sum- (f+g)(x)=f(x)+g(x)
Difference- (f-g)(x)=f(x)-g(x)
Product- (f*g)(x)=f(x)*g(x)
Quotient- (f/g)(x)=f(x)/g(x)
***REMEMBER THIS: f(x) is the notation when you use f(5) or f(y) 0r f(i^2). The notation means to plug what is in parenthesis into the equation instead of x.
EXAMPLE:
f(x)=6x+1 g(x)=3+x
(f+g)(x)=(6x+1)+(3+x)
Answer: 7x+4
This is the easiest way that I can teach you how to do this. It is just plugging into formulas.
Thanks and have a GREAT day. :)
Well since i was only at school monday, i don't know much. We learned domain and range.
3 types:
1. polynomial-an equation with no variables in the denominator.
domain is always (-infinity, + infinity)
range is (-infinity, + infinity) if odd.
2. square roots
domain: 1. set inside equal to 0 and solve for x.
2. set up a number line.
3. plug in numbers (use non-negative intervals)
range: [vertical shift, infinity)
**if square root of a number minus x^2 range= [0, +square root]
3. fractions
domain: 1. set bottom equal to 0 and solve for x.
2. set up intervals.
range: 1. take limit as x --> infinity
2. set up intervals
A FEW RULES:
-use brackets for everything EXCEPT asymtotes, +/- infinity, or circles.
--fractions always use parenthesis ( )
I'm pretty good at finding domain and range when i have a graph but when it gets to having to solve i have problems.
The using limit to find range is throwing me off?
All i know is when i get back to school i'll need major help!
i'm sure heather can help me :)
3 types:
1. polynomial-an equation with no variables in the denominator.
domain is always (-infinity, + infinity)
range is (-infinity, + infinity) if odd.
2. square roots
domain: 1. set inside equal to 0 and solve for x.
2. set up a number line.
3. plug in numbers (use non-negative intervals)
range: [vertical shift, infinity)
**if square root of a number minus x^2 range= [0, +square root]
3. fractions
domain: 1. set bottom equal to 0 and solve for x.
2. set up intervals.
range: 1. take limit as x --> infinity
2. set up intervals
A FEW RULES:
-use brackets for everything EXCEPT asymtotes, +/- infinity, or circles.
--fractions always use parenthesis ( )
I'm pretty good at finding domain and range when i have a graph but when it gets to having to solve i have problems.
The using limit to find range is throwing me off?
All i know is when i get back to school i'll need major help!
i'm sure heather can help me :)
4-1
4-1
In Advanced Math we’re stepping out of trig for a while and going back to Chapter 4. Section 4-1 is about finding domain and range. Domain is from left to right and range is from bottom to top. There are a few guidelines to follow:
1. Polynomial (an equation with no variables in the denominator)
Domain: ALWAYS (-infinity, infinity)
Range: (-infinity, infinity) if exponent is odd
2. Square roots
Domain: A). Set inside =0 and solve for x.
B). Set up a number line.
C). Plug in numbers, use the non-neg intervals.
Range: (vertical shift, infinity)
** If square root #-x^2 then range is [0, +/- square root #)
3. Fractions
Domain: A). Set bottom =0 and solve for x.
B). Set up intervals.
Range: A). Take limit as x – infinity.
B). Set up intervals.
** Always use () with fractions.
4. Absolute Value
Domain: (-infinity, infinity)
Range: [shift, infinity) if + and (-infinity, shift] if –.
EX 1: Find domain and range: 2x^3+3x^2-6x
A). This is a polynomial, so the domain is automatically (-infinity, infinity).
C). The exponent of 2x is 3, which is odd, so we know that the range will be (-infinity, infinity).
EX 2: Find the domain and range: 2/x-3.
A). This is a fraction so we know we’re going to have to set the bottom =0 to find domain.
B). x-3=0, so x=3
C). The domain will be (-infinity, 3)u(3, infinity)
D). Now to find the range, find the limit. This is going back to Chapter 13.
E). According to the limit rules, it’s 0.
F). So, the range is (-infinity, 0)u(0, infinity)
In Advanced Math we’re stepping out of trig for a while and going back to Chapter 4. Section 4-1 is about finding domain and range. Domain is from left to right and range is from bottom to top. There are a few guidelines to follow:
1. Polynomial (an equation with no variables in the denominator)
Domain: ALWAYS (-infinity, infinity)
Range: (-infinity, infinity) if exponent is odd
2. Square roots
Domain: A). Set inside =0 and solve for x.
B). Set up a number line.
C). Plug in numbers, use the non-neg intervals.
Range: (vertical shift, infinity)
** If square root #-x^2 then range is [0, +/- square root #)
3. Fractions
Domain: A). Set bottom =0 and solve for x.
B). Set up intervals.
Range: A). Take limit as x – infinity.
B). Set up intervals.
** Always use () with fractions.
4. Absolute Value
Domain: (-infinity, infinity)
Range: [shift, infinity) if + and (-infinity, shift] if –.
EX 1: Find domain and range: 2x^3+3x^2-6x
A). This is a polynomial, so the domain is automatically (-infinity, infinity).
C). The exponent of 2x is 3, which is odd, so we know that the range will be (-infinity, infinity).
EX 2: Find the domain and range: 2/x-3.
A). This is a fraction so we know we’re going to have to set the bottom =0 to find domain.
B). x-3=0, so x=3
C). The domain will be (-infinity, 3)u(3, infinity)
D). Now to find the range, find the limit. This is going back to Chapter 13.
E). According to the limit rules, it’s 0.
F). So, the range is (-infinity, 0)u(0, infinity)
Sunday, January 2, 2011
Exam review, 13-6
From the first thing we learned to the last thing we learned, its funny that the first and last are 2 of my favorites. I liked doing sigmas, it was easy and I like getting visual pictures of things.
A sigma is a series written in condensed form. There are all kinds of things you can do with sigmas but I like to keep it simple and find things in the sigma such as limits of summation, summand, and index.
Limits of Summation: the numbers in the sigma
Summand: is f(x) or y
index: is the variable
*Ill use Z as my sigma for now.
8
Z 5r
r= 1
Find the summand? 5r is the summand
What is the index?r
What are the limits of summation? 8 and 1
Evaluate the Sigma? 5+10+15+20+25+30+35+40=180; 5*1 + 5*2 + 5*3 + 5*4 + 5*5 + 5*6 + 5*7 + 5*8=180
A sigma is a series written in condensed form. There are all kinds of things you can do with sigmas but I like to keep it simple and find things in the sigma such as limits of summation, summand, and index.
Limits of Summation: the numbers in the sigma
Summand: is f(x) or y
index: is the variable
*Ill use Z as my sigma for now.
8
Z 5r
r= 1
Find the summand? 5r is the summand
What is the index?r
What are the limits of summation? 8 and 1
Evaluate the Sigma? 5+10+15+20+25+30+35+40=180; 5*1 + 5*2 + 5*3 + 5*4 + 5*5 + 5*6 + 5*7 + 5*8=180
7-4
Other than converting degrees to radians and vice versa, section 7-4 was another thing I liked to do. 7-4 has to do with finding refrence angles. All you have to do with this is follow the steps and your good to go. Which is the main reason I like it.
7-4
To find a refrence angle:
1) Find the Quadrant the angle is in.
2) Determine if the trig functions is positive or negative.
3) Subtract 180 from the angle until the absolute value of theta is between 0 and 90.
4) If it is a trig chart angle plug in. If not leave it or plug it into the calculator.
Ex. Find the refrence angle for sin675 degrees
675 degrees is in quadrant 4 where sin is negative.
675-180-180-180=45
sin45 is on the trig chart so your answer is square root of 2/2
7-4
To find a refrence angle:
1) Find the Quadrant the angle is in.
2) Determine if the trig functions is positive or negative.
3) Subtract 180 from the angle until the absolute value of theta is between 0 and 90.
4) If it is a trig chart angle plug in. If not leave it or plug it into the calculator.
Ex. Find the refrence angle for sin675 degrees
675 degrees is in quadrant 4 where sin is negative.
675-180-180-180=45
sin45 is on the trig chart so your answer is square root of 2/2
Trig Functions and Trig Chart
Another one of my favorite things that we did was memorizing the trig functions and memorizing the trig chart. These tasks had to do with memory and though they didn't seem like they would be easy it turned out not to be so bad.
Trig Funtions:
sin- y/r
cos- x/r
tan- y/x
cot- x/y
sec- r/x
csc- r/y
Trig Chart
In order of 0, pie/6, pie/4, pie/3, and pie/2
sin-0, 1/2, square root of 2/2, square root of 3/2 , and 1
cos- 1, square root of 3/2, square root of 2/2, 1/2, and 0
tan-0, square root of 3/3, 1, square root of 3, and Undefined
cot- Undefined, Square root of 3, 1, square root of 3/3 and 0
sec- 1, 2/square root of 3, square root of 2, 2, and Undefined
csc- Undefined, 2, square root of 2, 2/ square root of 3, and 1
Trig Funtions:
sin- y/r
cos- x/r
tan- y/x
cot- x/y
sec- r/x
csc- r/y
Trig Chart
In order of 0, pie/6, pie/4, pie/3, and pie/2
sin-0, 1/2, square root of 2/2, square root of 3/2 , and 1
cos- 1, square root of 3/2, square root of 2/2, 1/2, and 0
tan-0, square root of 3/3, 1, square root of 3, and Undefined
cot- Undefined, Square root of 3, 1, square root of 3/3 and 0
sec- 1, 2/square root of 3, square root of 2, 2, and Undefined
csc- Undefined, 2, square root of 2, 2/ square root of 3, and 1
7-1
7-1 has to be the easiest section we've ever learned. At this time I was thinking hmm maybe advanced math won't be that bad. Was I ever wrong. But anyway it was a great way to start.
The whole chapter 7 was pretty easy and I think we should go back and take some tests on it. Its been a very busy holiday for me, mainly because of basketball and I don't feel like I had much of a break. But it was all worth it. I got almost everything I wanted for christmas but I'm still waiting on the car. Anyway back to math.
7-1 converting degrees to radians
pie/180
38 degrees* Pie/180 = 38pie/180= 19pie/90
7-1 converting radians to degrees
180/pie
19pie/90 * 180/pie = 3420 pie/ 90pie= 38 degrees
The whole chapter 7 was pretty easy and I think we should go back and take some tests on it. Its been a very busy holiday for me, mainly because of basketball and I don't feel like I had much of a break. But it was all worth it. I got almost everything I wanted for christmas but I'm still waiting on the car. Anyway back to math.
7-1 converting degrees to radians
pie/180
38 degrees* Pie/180 = 38pie/180= 19pie/90
7-1 converting radians to degrees
180/pie
19pie/90 * 180/pie = 3420 pie/ 90pie= 38 degrees
So here is my third blog for the holidays. i decided to do it on the 6 trigonomic functions and what to do with them.
So the first one i am going to start with is tan. tan is opposite over adjacent. It can only be used when you have a right angle.
the second one i am going to explain is sine. Sine can only be used for right triangles too. It is opposite over hypotenuse. (hypotenuse is the side that is opposite of the right angle, by the way)
The third trig function i am going to talk about is cosine. Cosine is adjacent over hypotenuse. This one also cannot be used unless a right angle is present.
The fourth one i am going to talk about is csc. Csc is hypotenuse over opposite.
The fifth one is cot. Cot is adjacent over opposite.
The last one is sec. Sec is hypotenuse over adjacent.
With all of these, you can either find a side, or an angle. To find an angle just use the inverse of the function.
So the first one i am going to start with is tan. tan is opposite over adjacent. It can only be used when you have a right angle.
the second one i am going to explain is sine. Sine can only be used for right triangles too. It is opposite over hypotenuse. (hypotenuse is the side that is opposite of the right angle, by the way)
The third trig function i am going to talk about is cosine. Cosine is adjacent over hypotenuse. This one also cannot be used unless a right angle is present.
The fourth one i am going to talk about is csc. Csc is hypotenuse over opposite.
The fifth one is cot. Cot is adjacent over opposite.
The last one is sec. Sec is hypotenuse over adjacent.
With all of these, you can either find a side, or an angle. To find an angle just use the inverse of the function.
Well, here is my second blog for the holidays. Again, i forgot my binder at school so i can not really do anything recent. So i will just do something easy, just like everyone else. converting from radians to degrees, and degrees to radians.
First, you have to know the formulas.
The formula for converting degrees to radians is: # of degrees x pi / 180 degrees.
The forumla for converting radians to degrees is: Radians x 180 degrees / pi.
Example 1:
convert 5 pi to degrees.
1. 5 pi / 4 x 180 degrees / pi.
2. 225 degrees.
Example 2:
convert 70 degrees to radians.
1. 70 degrees x pi / 180 degrees.
2. 7 pi / 18.
It seems like everyone else was doing a blog on something simple like this, so i decided to do it too. I hate procrastinating, by the way. it makes my life complicated.
First, you have to know the formulas.
The formula for converting degrees to radians is: # of degrees x pi / 180 degrees.
The forumla for converting radians to degrees is: Radians x 180 degrees / pi.
Example 1:
convert 5 pi to degrees.
1. 5 pi / 4 x 180 degrees / pi.
2. 225 degrees.
Example 2:
convert 70 degrees to radians.
1. 70 degrees x pi / 180 degrees.
2. 7 pi / 18.
It seems like everyone else was doing a blog on something simple like this, so i decided to do it too. I hate procrastinating, by the way. it makes my life complicated.
Well, this is my first blog for the holidays. Maybe i shouldn’t have waited so long to do it, but oh well. Okay, i can’t do a blog about the last few thing we learned because i forgot my binder at school. But i will do it on the things i can remember; finding coterminal angles.
When in radians you either add or subtract two pi, depending whether or not they want a positive or negative angle. If they want a negative angle, subtract 2 pi. If they want a positive angle, add 2 pi.
Example 1: find a positive coterminal angle for pi / 2.
pi / 2 +2pi = 5pi / 2
When in degrees you either add or subtract 360 degrees. If they ask for a positive coterminal angle, then you add 360 degrees. If they want a negative coterminal angle, you subtract 360 degrees.
Example 2: find a negative coterminal angle for 32 degrees.
32 degrees – 360 degrees = -328 degrees.
Well, thats pretty much it.
When in radians you either add or subtract two pi, depending whether or not they want a positive or negative angle. If they want a negative angle, subtract 2 pi. If they want a positive angle, add 2 pi.
Example 1: find a positive coterminal angle for pi / 2.
pi / 2 +2pi = 5pi / 2
When in degrees you either add or subtract 360 degrees. If they ask for a positive coterminal angle, then you add 360 degrees. If they want a negative coterminal angle, you subtract 360 degrees.
Example 2: find a negative coterminal angle for 32 degrees.
32 degrees – 360 degrees = -328 degrees.
Well, thats pretty much it.
everyone else is doing it
Everyone else is making a post about converting radians to degrees and vice versa, so I want to jump on that bandwagon.
Ex. Convert 1.43pi to degrees.
Step1. 1.43pi x 180/pi
Step2. = 257.40 degrees
Ex. Convert 163 degrees to radians
Step1. 163 x pi/180
Step2. =163pi/180
Step3.=.9055
Ex. Convert 1.43pi to degrees.
Step1. 1.43pi x 180/pi
Step2. = 257.40 degrees
Ex. Convert 163 degrees to radians
Step1. 163 x pi/180
Step2. =163pi/180
Step3.=.9055
Holiday Blog 3
Holiday Blog 3
Last but not least here is my third and last blog of the holidays. Same as the other blogs, its been a fun break and not looking forward to school on Monday! So yall know I forgot my binder so all these problems are coming from off the top of my head. So since 7 is the easiest for me to remember I will be doing more problems from it.
Find a coterminal for each
Example 1:
139 degrees + 360 degrees = 499 degrees
Example 2:
12/26 + 2pi = 32/13
You either add or subtract 360 degrees or add or subtract 2pi to find a coterminal angle.
Okay so school starts tomorrow and it totally sucks. I am so not ready to go back. My only comfort is that we get 2 half days this week and the week after that we get a 4 day weekend!
Last but not least here is my third and last blog of the holidays. Same as the other blogs, its been a fun break and not looking forward to school on Monday! So yall know I forgot my binder so all these problems are coming from off the top of my head. So since 7 is the easiest for me to remember I will be doing more problems from it.
Find a coterminal for each
Example 1:
139 degrees + 360 degrees = 499 degrees
Example 2:
12/26 + 2pi = 32/13
You either add or subtract 360 degrees or add or subtract 2pi to find a coterminal angle.
Okay so school starts tomorrow and it totally sucks. I am so not ready to go back. My only comfort is that we get 2 half days this week and the week after that we get a 4 day weekend!
Promp 3
The thing I struggle with most in Algebra II is factoring equations. For some reason it doesn't stick in my head, which is a pain. Everything we do seems to always come back to factoring. I think it all goes back to letters don't belong in math. Maybe they're ok if the directions say what each one is equal to, but otherwise pluuuuuuuuhhhhh, thats the noise your tounge makes when you do that thing with it. Mrs. Brob it would be extremely helpful if we could spend maybe 10 minutes on doing Factoring. Otherwise, i'm cool with Algebra II.
Holiday Blog 2
Holiday Blog 2
Now it’s time for my second blog of the break. Once again hope everyone is enjoying break. It’s been a fun one for me but back to business. As I said in my last blog I forgot my binder at school so I am going to have to do some easy stuff off the top of my head.
This is converting minutes to degrees and minutes to seconds:
Example 1:
Convert 275 degrees
275 degrees + 60/60 + 80/3600 = 276.022
Example 2:
65. 71 x .60 = 42.6
.6 x .60 = .36
Sorry if some of this is wrong. I did the best I could without my resources. Everyone enjoy the rest of break and I will see you guys Monday!
Okay technically Monday is tomorrow. I wrote all these blogs early on in the break and I am now just posting them. Ehhhh school starts tomorrow ):
Now it’s time for my second blog of the break. Once again hope everyone is enjoying break. It’s been a fun one for me but back to business. As I said in my last blog I forgot my binder at school so I am going to have to do some easy stuff off the top of my head.
This is converting minutes to degrees and minutes to seconds:
Example 1:
Convert 275 degrees
275 degrees + 60/60 + 80/3600 = 276.022
Example 2:
65. 71 x .60 = 42.6
.6 x .60 = .36
Sorry if some of this is wrong. I did the best I could without my resources. Everyone enjoy the rest of break and I will see you guys Monday!
Okay technically Monday is tomorrow. I wrote all these blogs early on in the break and I am now just posting them. Ehhhh school starts tomorrow ):
inverses.
Chapter 8 had many sections that i found a little challenging. They were indeed doable of course but required a little extra attention. One of the more easier tasks of chapter eight was finding the inverse. These kind of problems were kind of fun too:)
Chapter *8 Finding the inverse
Ex.
3Cosx=1
reminders: calculator must be in the degree mode. Not the radian mode.
1.You have to get the equation by itself.
first divide “cosx’ by 3.
- 3cosx/3 = 1/3
2.You have to solve for x. x =the inverse of cosine.
X=Cos^-1(1/3)
3.You calculate (enter into your calculator) the inverse of cosine of (1/3).
Your answer should be 70.528 degrees. Round to the nearest tenth of a degree if not stated.
X=70.5°
Chapter *8 Finding the inverse
Ex.
3Cosx=1
reminders: calculator must be in the degree mode. Not the radian mode.
1.You have to get the equation by itself.
first divide “cosx’ by 3.
- 3cosx/3 = 1/3
2.You have to solve for x. x =the inverse of cosine.
X=Cos^-1(1/3)
3.You calculate (enter into your calculator) the inverse of cosine of (1/3).
Your answer should be 70.528 degrees. Round to the nearest tenth of a degree if not stated.
X=70.5°
#3
I hope everyone enjoyed their holidays but back to school tomorrow. Just a few more months you guys!
so for some reason I like to do the inscribe problems.
FORMULA:
a=nr^2 sin 360/2n cos 360/2n
EXAMPLE:
n=3 r=5in
3(5^2) sin 360/10 cos 360/10
*plug into your calculator!
your answer is well i wish i could tell you the answer but my batteries in my calculator are dead and we have none over here. I guess ill be going to the store in the morning before school.
I just have to add I HATE IDENTITIES.
I'm not sure what to talk about since we haven't learned anything new in the past few weeks or been at school. I'm ready to learn so i can have something to blog about.
so for some reason I like to do the inscribe problems.
FORMULA:
a=nr^2 sin 360/2n cos 360/2n
EXAMPLE:
n=3 r=5in
3(5^2) sin 360/10 cos 360/10
*plug into your calculator!
your answer is well i wish i could tell you the answer but my batteries in my calculator are dead and we have none over here. I guess ill be going to the store in the morning before school.
I just have to add I HATE IDENTITIES.
I'm not sure what to talk about since we haven't learned anything new in the past few weeks or been at school. I'm ready to learn so i can have something to blog about.
Polar.
HAPPY NEW YEARS!
I hope every had a safe and wondersful holiday:)Now its time to re-focus and get ready to begin school again. So here is holiday blog number two i'm sure.
What is Polar.
Chapter eleven is about polar and nonpolar. In section eleven one we learned conversions of polar to noon polar and vice versa. (x,y) coordinates are nonpolar. R and theta coordinates are polar.
X=rcostheta y=rsintheta
The same formula Is used to convert from polar to rectangular.
R=sqrtx^2+y^2
Give polar coordinates for (3,3)
R=sqrt 3^2=3^2
r=sqrt9+9
r=3sqrt2
tan x=3/3
x=tan-1(1) x=45 from trig chart
(3sqrt2,45 degrees)
I hope every had a safe and wondersful holiday:)Now its time to re-focus and get ready to begin school again. So here is holiday blog number two i'm sure.
What is Polar.
Chapter eleven is about polar and nonpolar. In section eleven one we learned conversions of polar to noon polar and vice versa. (x,y) coordinates are nonpolar. R and theta coordinates are polar.
X=rcostheta y=rsintheta
The same formula Is used to convert from polar to rectangular.
R=sqrtx^2+y^2
Give polar coordinates for (3,3)
R=sqrt 3^2=3^2
r=sqrt9+9
r=3sqrt2
tan x=3/3
x=tan-1(1) x=45 from trig chart
(3sqrt2,45 degrees)
#2
Lets take a few steps back to chapter 9! yayyyy.........NOT.
first thing we did was solving right triangles.
SOHCAHTOA
I still suck at these so I won't do any examples :)
sin=opposite/hypotenuse
cos=adjacent/hypotenuse
tan=opposite/adjacent
csc=hypotenuse/opposite
sec=hypotenuse/adjacent
cot=adjacent/opposite
FINDING AREA!
**when you have a right triangle it's super easy!
FORMULA:
1/2(b)(h)
non-right triangles are easy too!
FORMULA:
1/2(a)(b)sin(angle b/w)
LAW OF SINES
-only non-right triangles
-must have an angle and opposite leg value
FORMULA:
sin(angle)/opposite leg=sin(angle2)/opposite leg2
*cross multiply to solve!
LAW OF COSINES
*this one sucks!
-use this when you can't possibly do anything else.
FORMULA:
(opp leg)^2=(adj leg)^2+(adj leg)^2-2(adj leg)(adj leg)cos(angle b/w)
first thing we did was solving right triangles.
SOHCAHTOA
I still suck at these so I won't do any examples :)
sin=opposite/hypotenuse
cos=adjacent/hypotenuse
tan=opposite/adjacent
csc=hypotenuse/opposite
sec=hypotenuse/adjacent
cot=adjacent/opposite
FINDING AREA!
**when you have a right triangle it's super easy!
FORMULA:
1/2(b)(h)
non-right triangles are easy too!
FORMULA:
1/2(a)(b)sin(angle b/w)
LAW OF SINES
-only non-right triangles
-must have an angle and opposite leg value
FORMULA:
sin(angle)/opposite leg=sin(angle2)/opposite leg2
*cross multiply to solve!
LAW OF COSINES
*this one sucks!
-use this when you can't possibly do anything else.
FORMULA:
(opp leg)^2=(adj leg)^2+(adj leg)^2-2(adj leg)(adj leg)cos(angle b/w)
Holiday Blog 1
Holiday Blog 1
So I hope everyone is having a wonderful Christmas holiday, I know I am. This last week is kind of stressful though with me putting all my homework off until the last minute but don’t worry I’ll get it done. Anyway, I forgot my math binder in my locker at school so I’m going to have to blog about some easy stuff that I remember off the top of my head. Chapter seven is super easy stuff so I’m going to work a few problems.
Here are some example problems of the easy stuff you will ever do in advanced math.
Example 1
Convert 39 degrees to radians
39 degrees x π/180 = 13π/60
Example 2
Convert 13Ï€/45 to degrees
13Ï€/45 x 180/Ï€ = 2340/45 = 52
Hope everyone enjoys the rest of break! Happy Early New Year! See you guys on Monday January 3, 2011 (: Yayyyy!
So I hope everyone is having a wonderful Christmas holiday, I know I am. This last week is kind of stressful though with me putting all my homework off until the last minute but don’t worry I’ll get it done. Anyway, I forgot my math binder in my locker at school so I’m going to have to blog about some easy stuff that I remember off the top of my head. Chapter seven is super easy stuff so I’m going to work a few problems.
Here are some example problems of the easy stuff you will ever do in advanced math.
Example 1
Convert 39 degrees to radians
39 degrees x π/180 = 13π/60
Example 2
Convert 13Ï€/45 to degrees
13Ï€/45 x 180/Ï€ = 2340/45 = 52
Hope everyone enjoys the rest of break! Happy Early New Year! See you guys on Monday January 3, 2011 (: Yayyyy!
#1
Happy new year everyone! I actually miss some of you and can't wait to see ya'll tomorrow. I honestly hope I remember how to even do math since i've been lazy for the past 2 weeks :) Don't you just love last minute blogging? I dont know what to blog about anymore so i'll go back to chapter 7 (the easy chapter)
First week of school.
converting degrees to radians and radians to degrees.
DEGREES TO RADIANS.
multiply by pi/180
EXAMPLE:
15
15xpi/180=15pi.180
simplify!
your answer is pi/12
RADIANS TO DEGREES.
multiply by 180/pi
EXAMPLE:
3pi
3pix180/pi
pi's cancel out.
your answer is 540 degrees.
see how simple this stuff is!
First week of school.
converting degrees to radians and radians to degrees.
DEGREES TO RADIANS.
multiply by pi/180
EXAMPLE:
15
15xpi/180=15pi.180
simplify!
your answer is pi/12
RADIANS TO DEGREES.
multiply by 180/pi
EXAMPLE:
3pi
3pix180/pi
pi's cancel out.
your answer is 540 degrees.
see how simple this stuff is!
Chapter ten/ two
Chapter 10 section 2 is about plugging into one side of the formula when given an another one. They have many different formulas that you will need to learn in order to figure out the problems. Also you will need to know how to find the reference angle and also you will once again need to know your trig chart.
FORMULAS:
tan(alpha+beta)= tanalpha+tanbeta/1-tanalphatanbeta
tan(alpha- beta)= tanalpha-tanbeta/1+tanalphatanbeta
*cot is the reciprocal
i.e. cot(alpha+beta)=1-cotalphacotbeta/cotalpha+cotbeta
ect...
EXAMPLE:
simplify
tan90-tan60/1+tan90tan60
tan(90-60)
tan(30)
answer:squareroot(3)/3
This is how you do section ten dash two. have a great new year :)
FORMULAS:
tan(alpha+beta)= tanalpha+tanbeta/1-tanalphatanbeta
tan(alpha- beta)= tanalpha-tanbeta/1+tanalphatanbeta
*cot is the reciprocal
i.e. cot(alpha+beta)=1-cotalphacotbeta/cotalpha+cotbeta
ect...
EXAMPLE:
simplify
tan90-tan60/1+tan90tan60
tan(90-60)
tan(30)
answer:squareroot(3)/3
This is how you do section ten dash two. have a great new year :)
Chapter nine/ one
FORMULAS TO REMEMBER:
sin= opposite/hypotenuse sec= hypotenuse/adjacent
cos= adjacent/hypotenuse csc= hypotenuse/opposite
tan= opposite/adjacent cot= adjacent/opposite
EXAMPLE:
For triangle ABC
A= 90(degrees) B= 16 a= 34
First, since we already have A and B we need to find C. In order to do that all we are going to do is take 180 (which is the measure of a right triangle) minus 90 your right angle minus 16 the measure of angle B. Which will leave you with 74 and that is angle C.
Next, you have to find b and c. In order to do that you are going to have to take sin, cos, or tan of the angle B using your formula. For this problem you will use sin.
sin= opposite/ hypotnuse
sin 16 degrees= b/34
34sin16= 9.37
Now, you are going to need to find C. You are going to have to use cos= adjacent/ hypotnuse.
cos 16 degrees= c/34
34cos16= 32.68
And that is how you do section 9-1, have a great new year. :)
sin= opposite/hypotenuse sec= hypotenuse/adjacent
cos= adjacent/hypotenuse csc= hypotenuse/opposite
tan= opposite/adjacent cot= adjacent/opposite
EXAMPLE:
For triangle ABC
A= 90(degrees) B= 16 a= 34
First, since we already have A and B we need to find C. In order to do that all we are going to do is take 180 (which is the measure of a right triangle) minus 90 your right angle minus 16 the measure of angle B. Which will leave you with 74 and that is angle C.
Next, you have to find b and c. In order to do that you are going to have to take sin, cos, or tan of the angle B using your formula. For this problem you will use sin.
sin= opposite/ hypotnuse
sin 16 degrees= b/34
34sin16= 9.37
Now, you are going to need to find C. You are going to have to use cos= adjacent/ hypotnuse.
cos 16 degrees= c/34
34cos16= 32.68
And that is how you do section 9-1, have a great new year. :)
Sector of a circle
Formulas:
1. s=(r)(theta)
(s= arch length) (r= radius) (theta= central angle)
2. k= (1/2)(r^2)(s)
(k= area of the sector)
3. k= (1/2)(r)(s)
Next here are your steps:
1. find out what is given.. for example s=? r=? k=? and theta=?
You must find at least two of these in order to figure out this problem.
2. you are going to want to make sure that theta is in radians.
if theta is in degrees then you need to multiply that by (pie/180) to convert degrees into radians.
3. next, you are going to need to choose one of the formulas to plug into.
you want to choose the formula that works best with the problem.
for example if you have s= 4 cm and k= 36 cm you are not going to want to choose formula #1.
you would want to choose formula #3.
4. then, after you find one of the missing variables you will have one more variable still to find.
so, you are going to plug into another equation that fits the found variables and plug into the appropriate equation.
Here is an EXAMPLE:
A sector of a circle has an arc length of 4 cm and an area of 52 cm^2. Find its radius and the measure of its central angle.
So now follow the steps..
What are the given variables?
s= 4cm k=52cm^2 r=? theta=?
Now you must use the appropriate formula
formula #3 will work the best. [k=(1/2)(r)(s)]
52cm^2=(1/2)(r)(4)
52cm^2=2r
r= 26cm
Now that you found the answer for the variable r you can plug into another formula to find theta.
the best fit formula to find theta is formula #1 [s=(r)(theta)]
4cm= (26cm)(theta)
theta=(1/6 radians)
This is how you find the sector of the circle, and have a great new year.
1. s=(r)(theta)
(s= arch length) (r= radius) (theta= central angle)
2. k= (1/2)(r^2)(s)
(k= area of the sector)
3. k= (1/2)(r)(s)
Next here are your steps:
1. find out what is given.. for example s=? r=? k=? and theta=?
You must find at least two of these in order to figure out this problem.
2. you are going to want to make sure that theta is in radians.
if theta is in degrees then you need to multiply that by (pie/180) to convert degrees into radians.
3. next, you are going to need to choose one of the formulas to plug into.
you want to choose the formula that works best with the problem.
for example if you have s= 4 cm and k= 36 cm you are not going to want to choose formula #1.
you would want to choose formula #3.
4. then, after you find one of the missing variables you will have one more variable still to find.
so, you are going to plug into another equation that fits the found variables and plug into the appropriate equation.
Here is an EXAMPLE:
A sector of a circle has an arc length of 4 cm and an area of 52 cm^2. Find its radius and the measure of its central angle.
So now follow the steps..
What are the given variables?
s= 4cm k=52cm^2 r=? theta=?
Now you must use the appropriate formula
formula #3 will work the best. [k=(1/2)(r)(s)]
52cm^2=(1/2)(r)(4)
52cm^2=2r
r= 26cm
Now that you found the answer for the variable r you can plug into another formula to find theta.
the best fit formula to find theta is formula #1 [s=(r)(theta)]
4cm= (26cm)(theta)
theta=(1/6 radians)
This is how you find the sector of the circle, and have a great new year.
Lovin' some math over my holidays.
So, as we all know i do not remember anything that is going on with school and i am pretty sure i am not the only one. I see everyone posting blogs about what we learned in certain chapters, but i will be fair and not copy what they are putting. I will also confess that i do not have my binder therefore i cant exactly say anything we had learned. Actually, this may not even count as a blog because there is nothing to do with math in this but atleast i still came to this website. Well, like some other people i can openly say that I would not be surprised if i failed the midterm. I guess we will have to see when the grades come back, BUT my luck..my family decides i will leave for a 8 day vacation the day of the hardest exam ever! I do know that school comes first, but i honestly packed instead.. I did read over everything for the exam but it just doesn't click to me. Don't get me wrong i do try, i just will never understand. Well i'm gonna stop exposing myself now and check out those prompts i didn't do yet..
Happy New Year! :)
Happy New Year! :)
Saturday, January 1, 2011
13-6
13-6 is all about sigma. A sigma is a series written in condensed form. The sigma is representing with F because I’m too lazy to make it.
The Sigma contains three parts. The summand is the limits of summation, and the index. The number to the right of the Sigma is the summand. The number on top of the Sigma is the limits of summation. The number on the bottom of the Sigma is the index. You can either be asked to expand the Sigma or evaluate the sigma. To expand you only plug the numbers was they need to go. To evaluate you solve the whole thing.
The greek letter sigma is often used in mathematics to express a series or its sum in abbreviated form.
6
F4k
K = 2
Find the Summand? 4k
What is the index? k
What are the limits of summation? 5 and 2
Evaluate the Sigma
4(2) + 4(3) + 4(4) + 4(5) + 4(6)
8 + 12 + 16+ 20 + 24 = 80
Basically 13-6 is easy but it can mess you over if you don't know all the information about stigma if you do you will easily breeze threw this section.
The Sigma contains three parts. The summand is the limits of summation, and the index. The number to the right of the Sigma is the summand. The number on top of the Sigma is the limits of summation. The number on the bottom of the Sigma is the index. You can either be asked to expand the Sigma or evaluate the sigma. To expand you only plug the numbers was they need to go. To evaluate you solve the whole thing.
The greek letter sigma is often used in mathematics to express a series or its sum in abbreviated form.
6
F4k
K = 2
Find the Summand? 4k
What is the index? k
What are the limits of summation? 5 and 2
Evaluate the Sigma
4(2) + 4(3) + 4(4) + 4(5) + 4(6)
8 + 12 + 16+ 20 + 24 = 80
Basically 13-6 is easy but it can mess you over if you don't know all the information about stigma if you do you will easily breeze threw this section.
11-2
In chapter 11 section 2 it is dealing wit complex numbers.
Rectangular= -2=x+yi
Polar= -2=rcostheta+rsinthetai-> from 11-1
If u want to abbreviate to 2=rcis
To multiply complex numbers
Rectangular – foil
Polar – multiply r; add thetas
Express
2cis45degrees to rectangular
x=rcostheta=2cos45degrees=2(square root of 2/2)=square root of 2
y=rsintheta=2sin45degrees=2(square root of 2/2)=square root of 2
z=square root of 2+square root of 2 i
Express 2
Z=1-square root of 3 in polar
Square root of 1^2 +square root of 3^2=1+3= square root of 4 = +/-2
Tan –square root of 3/1=theta=tan^-1square root of 3
60 where it – second Q and 3rd .
Q2= 120degrees Q3= 300 degrees.
Z=2cis300degrees
Z=2cos300degrees+2sin300degreesi
Z=-2cis120degrees
Z=-2cos 120degrees+-2sin 120degreesi
All you answer above
Z1*Z2
Z1 2cis 30 z2 3 cis 20
= 6cis 50degrees
Basically section11-2 is weird long and aggravating. If you know what you are doing and know all your formulas it will be easy for you
Rectangular= -2=x+yi
Polar= -2=rcostheta+rsinthetai-> from 11-1
If u want to abbreviate to 2=rcis
To multiply complex numbers
Rectangular – foil
Polar – multiply r; add thetas
Express
2cis45degrees to rectangular
x=rcostheta=2cos45degrees=2(square root of 2/2)=square root of 2
y=rsintheta=2sin45degrees=2(square root of 2/2)=square root of 2
z=square root of 2+square root of 2 i
Express 2
Z=1-square root of 3 in polar
Square root of 1^2 +square root of 3^2=1+3= square root of 4 = +/-2
Tan –square root of 3/1=theta=tan^-1square root of 3
60 where it – second Q and 3rd .
Q2= 120degrees Q3= 300 degrees.
Z=2cis300degrees
Z=2cos300degrees+2sin300degreesi
Z=-2cis120degrees
Z=-2cos 120degrees+-2sin 120degreesi
All you answer above
Z1*Z2
Z1 2cis 30 z2 3 cis 20
= 6cis 50degrees
Basically section11-2 is weird long and aggravating. If you know what you are doing and know all your formulas it will be easy for you
9-3
In chapter 9 section 3 it is dealing with the law of sines. The law of sines has 2 steps to follow: used with non-right triangles, only use when you have an angle and opp leg value you know.
The formula for law of sine is sin(angle)/opp leg=sin(angle 2)/ opp leg 2
*cross multiply to solve
Example is in triangle rst if so find all possible measure of
Sin 126^o/12* sin t/7=12sint=7sin126= sin t= 7 sin 126degrees/12
You get 28.159degrees which is your answer.
Example 2:
Solve for x if one angle is 60 degrees another 25 degrees and a side is 8.
Sin 60 degrees/8=sin 25degrees/x = xsin60degrees=8sin25degrees you divide sin 60 degrees you get 3.904 as your answer.
Basically law of sine can be tricky at times but if you on a role and know everything you will make law of sine your pet and own at it.
The formula for law of sine is sin(angle)/opp leg=sin(angle 2)/ opp leg 2
*cross multiply to solve
Example is in triangle rst
Sin 126^o/12* sin t/7=12sint=7sin126= sin t= 7 sin 126degrees/12
You get 28.159degrees which is your answer.
Example 2:
Solve for x if one angle is 60 degrees another 25 degrees and a side is 8.
Sin 60 degrees/8=sin 25degrees/x = xsin60degrees=8sin25degrees you divide sin 60 degrees you get 3.904 as your answer.
Basically law of sine can be tricky at times but if you on a role and know everything you will make law of sine your pet and own at it.
8-1
In Section 8-1 to solve for theta you get the trig function by itself and then take an inverse.
An inverse has 2 answers with some exceptions; find where the angle is based on trig function and if number +ve or –ve.
Steps: take the inverse of +ve number to find the Q1 angle.
To get to
Q2 make –ve degree add 180
Q3 add 180 degrees
Q4 make +ve add 360 degrees
Example:
3 cosine(theta)= 1
1. Subtract 3 from one, which will equal -1/3
2. Then do the cosine inverse(1/3)=
70.529
3. We are looking for the positive cosine, which is x.
4. 70.529 degrees is in the first quadrant, so it is positive.
5. Since it is in the first and fourth quadrant, for fourth quadrant, we will do 360-70.529, and it will equate to 289.471 degrees
Answers: 70.529(degrees), 289.471(degrees).
If u know everything in 8-1 and the formulas u will do fine and it will be a breeze.
An inverse has 2 answers with some exceptions; find where the angle is based on trig function and if number +ve or –ve.
Steps: take the inverse of +ve number to find the Q1 angle.
To get to
Q2 make –ve degree add 180
Q3 add 180 degrees
Q4 make +ve add 360 degrees
Example:
3 cosine(theta)= 1
1. Subtract 3 from one, which will equal -1/3
2. Then do the cosine inverse(1/3)=
70.529
3. We are looking for the positive cosine, which is x.
4. 70.529 degrees is in the first quadrant, so it is positive.
5. Since it is in the first and fourth quadrant, for fourth quadrant, we will do 360-70.529, and it will equate to 289.471 degrees
Answers: 70.529(degrees), 289.471(degrees).
If u know everything in 8-1 and the formulas u will do fine and it will be a breeze.
The confusing Chapter 13
Happy New Years everyone! Chapter 13 is still pretty hard for me being I never really understood the concept. In this particular section I was able to clearly understand what it was telling you to do, but when I had to plug into the calculator I struggled to tell whether it was reaching 1 or 0. Besides knowing where the numbers were aiming to, you needed to remember all of the guidelines which sometimes where hard to not mix up.
Guidlines for this Chapter:
For Fractions:
1. If the top degree = bottom degree the answer is a coefficient.
2. If the top degree > bottom degree the answer is +/- infinity.
3. If the top degree < bottom degree the answer is 0. If the rules above do not apply to the given, then you simply use the table function in your calculator.
For example:
1. Lim/n->infinity n^7 +6/ 9n^2 – 7n
The degrees are 7 and 5.
The top one (7) is larger than the bottom one (2).
When you look at your rules, they state that the answer will be +/- infinity.
Happy New Years once again and see yall Monday (:
Guidlines for this Chapter:
For Fractions:
1. If the top degree = bottom degree the answer is a coefficient.
2. If the top degree > bottom degree the answer is +/- infinity.
3. If the top degree < bottom degree the answer is 0. If the rules above do not apply to the given, then you simply use the table function in your calculator.
For example:
1. Lim/n->infinity n^7 +6/ 9n^2 – 7n
The degrees are 7 and 5.
The top one (7) is larger than the bottom one (2).
When you look at your rules, they state that the answer will be +/- infinity.
Happy New Years once again and see yall Monday (:
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