I totally forgot about this. I've been doing that Aleks thing all day and you know how time filies when you're having too much fun.
Anyway, like everyone has said before, we learned about Matrices and Cramer's Rule. But I think anyone convered transpose, so I go over that.
Transpose is really simple, just switch the dimensions of the matrix. ex. 2x3 = 3x2
[2 3
5 6
9 1 ]
= [ 2 5 9
3 6 1]
There you go.
Monday, January 31, 2011
Sunday, January 30, 2011
maaaaarrrrrryy
matrices, i learned alot between this class and number theory so i'm just gonna go over what i know because i can't remember what i learned in which class.
First of all.
when you add matrices, you just add the corresponding numbers, if you need an example, shame on you.
when you subtract, it goes the same as adding, your answer will be the same dimensions as the two you are adding/subtracting because they have to be the same dimensions to perform those actions anyway
Ex.3 4 - 4 6 = -1 -2
2 0 5 8 -3 -8
Next.
Multiplying/dividing
when you multiply matrices, you first have to seee if you can by checking the dimensions.
Ex. 2 x 3 3 x 4 Possible
3 x 4 3 x 4 not possible
the inside numbers will be the same if its possible, and your product will be the outside numbers.
Ex. 2 x 3 x 3 x 4 = 2x4
and you multiply row by column, and you add numbers and stuff and its really complicated. but you can figure it out if you do row by column.
First of all.
when you add matrices, you just add the corresponding numbers, if you need an example, shame on you.
when you subtract, it goes the same as adding, your answer will be the same dimensions as the two you are adding/subtracting because they have to be the same dimensions to perform those actions anyway
Ex.3 4 - 4 6 = -1 -2
2 0 5 8 -3 -8
Next.
Multiplying/dividing
when you multiply matrices, you first have to seee if you can by checking the dimensions.
Ex. 2 x 3 3 x 4 Possible
3 x 4 3 x 4 not possible
the inside numbers will be the same if its possible, and your product will be the outside numbers.
Ex. 2 x 3 x 3 x 4 = 2x4
and you multiply row by column, and you add numbers and stuff and its really complicated. but you can figure it out if you do row by column.
kaitlyn's bloggyyy
This weekk we went over matrices yayy! it was either chapter 4 or 14, i dont remember. matrices are probably the easiest thing we have done so far this year.
When you add two matrices together, you just add the two numbers that line up with each other in the two matrices. the same goes for subtracting.
Ex: [2 4] + [6 3]= [8 7]
[5 2] [8 1] [13 3]
When you multiply matrices, you have to check the dimensions first. Take for example, if you have a 2x3 matrix, and a 3x1, you can multiply these because the two inside numbers are the same. if you have a 2x1 and a 5x2, you can't multiply these.
Ex: [2 1] x [1 0]= [2+2 0+2] = [4 2]
[3 5] [2 2] [3+10 0+10] [13 10]
so that is pretty much the overall of matrices, i like matrices the most in math. they are kinda cool, i guesssssss. yeahh so thats all(:
When you add two matrices together, you just add the two numbers that line up with each other in the two matrices. the same goes for subtracting.
Ex: [2 4] + [6 3]= [8 7]
[5 2] [8 1] [13 3]
When you multiply matrices, you have to check the dimensions first. Take for example, if you have a 2x3 matrix, and a 3x1, you can multiply these because the two inside numbers are the same. if you have a 2x1 and a 5x2, you can't multiply these.
Ex: [2 1] x [1 0]= [2+2 0+2] = [4 2]
[3 5] [2 2] [3+10 0+10] [13 10]
so that is pretty much the overall of matrices, i like matrices the most in math. they are kinda cool, i guesssssss. yeahh so thats all(:
Teeorees blog thingy.
matrices matrices matrices.. oh how i didnt miss you!
this week was a review of multiplying/dividing/adding/subtracting matrices.
it was easy but it was aggravating because i knew how to do it. however, this does not mean that i want to learn harder stuff because then i wont bother to learn it because i wont feel like bothering with it. oh welll onn to the explanationss!
When you are multiplying matrices, the dimensions of both matrices have to have the same numbers in the middle. Also when multiplying matrices, you go row by column.
Ex. 2 x 2 2 x 2
If those two numbers match, then the two outer numbers will give you what the dimensions of the final matrix will be.
Ex. 3 x 2 2 x 3 - Now you will end up with a 3x3 matrix.
We also learned about Cramer's rule:
D stands for the determinant.
dx/d
dy/d
dz/d if there is one.
kaybye!
this week was a review of multiplying/dividing/adding/subtracting matrices.
it was easy but it was aggravating because i knew how to do it. however, this does not mean that i want to learn harder stuff because then i wont bother to learn it because i wont feel like bothering with it. oh welll onn to the explanationss!
When you are multiplying matrices, the dimensions of both matrices have to have the same numbers in the middle. Also when multiplying matrices, you go row by column.
Ex. 2 x 2 2 x 2
If those two numbers match, then the two outer numbers will give you what the dimensions of the final matrix will be.
Ex. 3 x 2 2 x 3 - Now you will end up with a 3x3 matrix.
We also learned about Cramer's rule:
D stands for the determinant.
dx/d
dy/d
dz/d if there is one.
kaybye!
:D:D yay bloggy blogged blog by dylan
We learned about matrices. We learned how to add, subtract, multiply, and find the inverse and/or determinate. To add matrices, you have to have same dimension matrices, ie: 2x2 + 2x2. (subtraction follow this also) Multiplication requires the last number of the first dimension, 2x3, and the first number of the second dimension, 3x2, to be the same.
To find inverses: the inverse of A = 1/|A| * [d -b(first row) -c a(second row)]
For determinates: ad-bc
That's pretty much it :D
To find inverses: the inverse of A = 1/|A| * [d -b(first row) -c a(second row)]
For determinates: ad-bc
That's pretty much it :D
Nathan's blog
We have been going over algebra 2 stuff in math and i have to say it is pretty easy. This week we learned about matrices, which is beyond easy. We also started the online math work.
So, when you are multiplying matrices, the dimensions of both matrices have to have the same numbers in the middle. Also when multiplying matrices, you go row by column. Ex. 2 x 2 2 x 2
If those two numbers match, then the two outer numbers will give you what the dimensions of the final matrix will be. Ex. 3 x 2 2 x 3 - Now you will end up with a 3x3 matrix.
We also learned about Cramer's rule: I think that d stands for the determinant or something like that.
dx/d
dy/d
dz/d
So, overall, this algebra 2 review has been pretty easy. The matrix chapter is probably the easiest one that we have done all year.
So, when you are multiplying matrices, the dimensions of both matrices have to have the same numbers in the middle. Also when multiplying matrices, you go row by column. Ex. 2 x 2 2 x 2
If those two numbers match, then the two outer numbers will give you what the dimensions of the final matrix will be. Ex. 3 x 2 2 x 3 - Now you will end up with a 3x3 matrix.
We also learned about Cramer's rule: I think that d stands for the determinant or something like that.
dx/d
dy/d
dz/d
So, overall, this algebra 2 review has been pretty easy. The matrix chapter is probably the easiest one that we have done all year.
Lawrence's Blog
this week we learned about matrices. it is a total review from algebra 2. this section was very easy and i remembered it from last year. we learned all over again how to add and subtract which was the first section and the second section was multiplying the matrices. when you multiply matrices you do row by column. also when you do matrices you have to make sure they match. i give you a few examples.
2x1, 1x4 you can add, subtract, and multiply these because the ones are the same number.
4x2, 5x1 you cant do anything with this one because the 5 and the 2 are different numbers and it just wont workout.
i hope this helped a few people out with matrices and if you need help dont be afraid to ask me. i actually know what i am doing on these
2x1, 1x4 you can add, subtract, and multiply these because the ones are the same number.
4x2, 5x1 you cant do anything with this one because the 5 and the 2 are different numbers and it just wont workout.
i hope this helped a few people out with matrices and if you need help dont be afraid to ask me. i actually know what i am doing on these
Friday, January 28, 2011
Taylor's 7th (I think XD) Review Blog
REVIEW TIME!!
The following information is a review based on section 7.2 of my Advanced Math notes.
These are the formulas that you will need to work the following problems and equations.
S=RTheta(This is a circle with a line going horizontally through it)
k= 1/2r^2theta
k= 1/2rs
S stands for the arc length of the equation/formula.
R stands for the radius of the equation/formula.
Theta stands for the central angle of a problem.
K stands for the area of a sector.
Yet, sometimes you will not just be ask to find the arc length . They will sometimes call s, r, and
Theta other things in some word problems.
Theta is sometimes called the apparent size.
S is sometimes called the diameter of the object.
R is some times called the distance between the objects.
IMPORTANT NOTE THETA MUST BE IN RADIA NOT DEGREES!!
Now lets apply what we just learned into the following problems.
The arc length of an object is 4cm and the central angle is .0098 radians. Find the area and the radius of the object.
Theta= .0098
S= 4 cm
K= ?
R= ?
First identify what you know from the problem and what you need to find out.
4=Rx.0098
Next, chose one of the formulas and fill in what you can. Lets find R first since it is easier to find. Then fill in the appropriate places with the right numbers.
408.163 cm=R
Then divide .0098 from both sides to find out what R is. Make sure to put the length of measurements in the appropriate places and to round to the third place after the decimal.
K=1/2x408.163x4
Now lets use one of the K equations to find the area of the object. So once more fill in the appropriate areas with the right numbers.
K=816.326 cm^2
Finally, you just need to multiply the appropriate numbers and this is what you should get as your answer.
Ok lets try this problem next.
The distance between the moon and Earth is 5x10^7 mi and the apparent size of the moon is 2000 radians. What is the diameter?
Theta=2000 radians
S= ?
R= 5x10^7 mi
First, write down what you do know from the problem and what you need to find.
S=5x10^7x2000
Then place the information into a problem so that you can work it out.
S= 100000000000 mi
Then you multiply 5x10^7 and 2000 and you get your answer.
That is what we learned in lesson 7.2.
The following information is a review based on section 7.2 of my Advanced Math notes.
These are the formulas that you will need to work the following problems and equations.
S=RTheta(This is a circle with a line going horizontally through it)
k= 1/2r^2theta
k= 1/2rs
S stands for the arc length of the equation/formula.
R stands for the radius of the equation/formula.
Theta stands for the central angle of a problem.
K stands for the area of a sector.
Yet, sometimes you will not just be ask to find the arc length . They will sometimes call s, r, and
Theta other things in some word problems.
Theta is sometimes called the apparent size.
S is sometimes called the diameter of the object.
R is some times called the distance between the objects.
IMPORTANT NOTE THETA MUST BE IN RADIA NOT DEGREES!!
Now lets apply what we just learned into the following problems.
The arc length of an object is 4cm and the central angle is .0098 radians. Find the area and the radius of the object.
Theta= .0098
S= 4 cm
K= ?
R= ?
First identify what you know from the problem and what you need to find out.
4=Rx.0098
Next, chose one of the formulas and fill in what you can. Lets find R first since it is easier to find. Then fill in the appropriate places with the right numbers.
408.163 cm=R
Then divide .0098 from both sides to find out what R is. Make sure to put the length of measurements in the appropriate places and to round to the third place after the decimal.
K=1/2x408.163x4
Now lets use one of the K equations to find the area of the object. So once more fill in the appropriate areas with the right numbers.
K=816.326 cm^2
Finally, you just need to multiply the appropriate numbers and this is what you should get as your answer.
Ok lets try this problem next.
The distance between the moon and Earth is 5x10^7 mi and the apparent size of the moon is 2000 radians. What is the diameter?
Theta=2000 radians
S= ?
R= 5x10^7 mi
First, write down what you do know from the problem and what you need to find.
S=5x10^7x2000
Then place the information into a problem so that you can work it out.
S= 100000000000 mi
Then you multiply 5x10^7 and 2000 and you get your answer.
That is what we learned in lesson 7.2.
Charlie's (Jan28)
This week we did matrices stuff.
we added, subtracted, multiplied, & divided them.
(we started doing the ALEKS.com thing online)
we did some cramer guy's rule too.
cramers rule:
dx/d = x
dy/d = y
dz/d = z
dx [answer y z ...]
dy [x answer z ...]
d [x y z ...]
we also learned how to do inverses.
to find the inverse:
A^-1 = 1/the determinant of A
to find the determinant:
-[a b] = ad - bc
[c d]
Example ~inverse~:
1) find the inverse of [3 2]
_______________[1 6]
3 x 6 - 2 x 1 = 18 - 2 = 16
1/16
so the inverse of [3 2]
____________[1 6] is 1/16
we added, subtracted, multiplied, & divided them.
(we started doing the ALEKS.com thing online)
we did some cramer guy's rule too.
cramers rule:
dx/d = x
dy/d = y
dz/d = z
dx [answer y z ...]
dy [x answer z ...]
d [x y z ...]
we also learned how to do inverses.
to find the inverse:
A^-1 = 1/the determinant of A
to find the determinant:
-[a b] = ad - bc
[c d]
Example ~inverse~:
1) find the inverse of [3 2]
_______________[1 6]
3 x 6 - 2 x 1 = 18 - 2 = 16
1/16
so the inverse of [3 2]
____________[1 6] is 1/16
Monday, January 24, 2011
Feroz's Blog
Just woke up from a 10 hour nap. Let's get started.
Last week was a bit vague for me. Monday was MLK day, I missed Tuesday, we had a quiz Thursday, and I fell asleep during our test Friday. All in all, I don't know what I'm supposed to be covering. I guess functions?
From the little I understand, I know we have plug things in and stuff. So I'll start there.
ex. Find F(x) = 2x + 5
F(5) = 2(5) + 5
= 15
And a function within a function.
ex. Find (f o g)(x) if F(x) = 2x + 5 and g(x) = 3x + 2
= 2(3x+2) + 5
= 6x + 4 + 5
= 6x = -9
= x = -3/2
Alright, well that's it for what I understand. New semester and I'm still clueless. So much for trying.
Last week was a bit vague for me. Monday was MLK day, I missed Tuesday, we had a quiz Thursday, and I fell asleep during our test Friday. All in all, I don't know what I'm supposed to be covering. I guess functions?
From the little I understand, I know we have plug things in and stuff. So I'll start there.
ex. Find F(x) = 2x + 5
F(5) = 2(5) + 5
= 15
And a function within a function.
ex. Find (f o g)(x) if F(x) = 2x + 5 and g(x) = 3x + 2
= 2(3x+2) + 5
= 6x + 4 + 5
= 6x = -9
= x = -3/2
Alright, well that's it for what I understand. New semester and I'm still clueless. So much for trying.
Sunday, January 23, 2011
Tori's
In this blog, I will review section 4-2 which deals with notations.
f(x) is notation.
When f(5) or f(y) or f(i^2) the notation means to plug what is in the parenthesis into the equation instead of x.
1.) The sum of f and g is: (f+g)(x)
2.) The difference of f and g is: (f-g)(x)
3.) The product of f and g is: (f x g)(x)
4.) The quotient of f and g is:(f/g)(x)
Composition Functions - Function inside of another function.
(f o g)(x)
(g o f)(x)
Find f(5) if f(x) equals 3x+7
3(5)+7=22
Find (g o f)(x) if f(x)=5x+1 and g(x)=x/-1
(5x+1)/-1= -5x-1
f(x) is notation.
When f(5) or f(y) or f(i^2) the notation means to plug what is in the parenthesis into the equation instead of x.
1.) The sum of f and g is: (f+g)(x)
2.) The difference of f and g is: (f-g)(x)
3.) The product of f and g is: (f x g)(x)
4.) The quotient of f and g is:(f/g)(x)
Composition Functions - Function inside of another function.
(f o g)(x)
(g o f)(x)
Find f(5) if f(x) equals 3x+7
3(5)+7=22
Find (g o f)(x) if f(x)=5x+1 and g(x)=x/-1
(5x+1)/-1= -5x-1
lawrence's blog
for this blof i will go over section 3. i finally figured out how to do this part and now i find it very easy and hopefully i can keep figuring out how to do all the other stuff so i can pass the test. there are only a few formulas to follow and they are very easy.
1.) Y-axis - You have to plug in (-x)
2.) X-axis - You have to plug in (-y), then simplifiy; if the equations are equal then it has symmetry.
3.) Origin - You have to plug in (-x) and (-y)
4.) Y=X - You have to switch x and y, and then solve for y.
Section 4-1 dealt with domain and range, and finding the zeros. this was also a easy part that i found in chapter 4 and hopefully the written part has alot of these so i can pass that one too.
ex.
y=4x^2+6x-7
D:(-∞,∞)
R:
Zeros:(x+7)(x-1)
x=-7,1
1.) Y-axis - You have to plug in (-x)
2.) X-axis - You have to plug in (-y), then simplifiy; if the equations are equal then it has symmetry.
3.) Origin - You have to plug in (-x) and (-y)
4.) Y=X - You have to switch x and y, and then solve for y.
Section 4-1 dealt with domain and range, and finding the zeros. this was also a easy part that i found in chapter 4 and hopefully the written part has alot of these so i can pass that one too.
ex.
y=4x^2+6x-7
D:(-∞,∞)
R:
Zeros:(x+7)(x-1)
x=-7,1
Kaitlyn's blogggg
In this blogg, i will go over section 4-2. This section deals with notation. It was pretty easy for me because all you have to do is solve equations.
f(x) means notation. If you replace the x with a number, then you plug that number into the x values in the equation that follows it.
Sometime you have two different sets of equations that you might have to either add, subtract, multiply, or divide together: 1) (f+g)(x)
2) (f-g)(x)
3) (fxg)(x)
4) (f/g)(x)
Sometimes the functions will have to be put inside other functions: 1) f(g(x))
2) g(f(x))
Examples: f(x)=2x+4 find f(5)
-f(5)=2(5)+4
- f(5)=10+4
-f(5)=14 <---answerrrrr
Examples: f(x)=x+1 g(x)=2x-1 find (f+g)(x)
-(x+1)+(2x-1)
-(3x) <---answerrr
Examples: f(x)=x-1 g(x)=x^2+5 find g(f(x))
-(x-1)^2+5
-x^2-2x+1+5
-x^2-2x+6 <---answerrrr
f(x) means notation. If you replace the x with a number, then you plug that number into the x values in the equation that follows it.
Sometime you have two different sets of equations that you might have to either add, subtract, multiply, or divide together: 1) (f+g)(x)
2) (f-g)(x)
3) (fxg)(x)
4) (f/g)(x)
Sometimes the functions will have to be put inside other functions: 1) f(g(x))
2) g(f(x))
Examples: f(x)=2x+4 find f(5)
-f(5)=2(5)+4
- f(5)=10+4
-f(5)=14 <---answerrrrr
Examples: f(x)=x+1 g(x)=2x-1 find (f+g)(x)
-(x+1)+(2x-1)
-(3x) <---answerrr
Examples: f(x)=x-1 g(x)=x^2+5 find g(f(x))
-(x-1)^2+5
-x^2-2x+1+5
-x^2-2x+6 <---answerrrr
Nathan's blog
In this blog, I will review section 4-2 which deals with notations. I actually know this stuff because I made a 90 on the test on Friday.
f(x) is notation. When f(5) or f(y) or f(i^2) the notation means to plug what is in the parenthesis into the equation instead of x.
1.) The sum of f and g is: (f+g)(x)
2.) The difference of f and g is: (f-g)(x)
3.) The product of f and g is: (f x g)(x)
4.) The quotient of f and g is:(f/g)(x)
Composition Functions - Function inside of another function.
(f o g)(x)
(g o f)(x)
Find f(5) if f(x) equals 3x+7
3(5)+7=22
Find (f o g)(x) if f(x)=2x+5 and g(x)=3x+2
2(3x+2)+5
6x+4+5
6x=-9
x=-3/2
Overall I think that this was the easiest section of chapter 4. Chapter 4 wasn't really that hard, we just had to remember the algebra 2 stuff. So, now I am going to study for Monday's test.
f(x) is notation. When f(5) or f(y) or f(i^2) the notation means to plug what is in the parenthesis into the equation instead of x.
1.) The sum of f and g is: (f+g)(x)
2.) The difference of f and g is: (f-g)(x)
3.) The product of f and g is: (f x g)(x)
4.) The quotient of f and g is:(f/g)(x)
Composition Functions - Function inside of another function.
(f o g)(x)
(g o f)(x)
Find f(5) if f(x) equals 3x+7
3(5)+7=22
Find (f o g)(x) if f(x)=2x+5 and g(x)=3x+2
2(3x+2)+5
6x+4+5
6x=-9
x=-3/2
Overall I think that this was the easiest section of chapter 4. Chapter 4 wasn't really that hard, we just had to remember the algebra 2 stuff. So, now I am going to study for Monday's test.
maryblogblogblog
This week was all about inverses and graphing stuff.
These are the rules to see if something is symmetric
( i left my math binder at school so forgive me if i'm wrong)
to see if its symmetric on the x axis, you plug in negative y-and see if this equation and the original match
to see if its symmetric on the y axis, you plug in negative x-ditto
to see if its symmetric on the origin, you plug in negative x and y-ditto
to see if its symmetric ont the y=x, you switch the x and ys.-and ditto
example:
f(x)= x^2+xy
x axis-x^2-xy-not the same equation so noooooo
y axis-(-X^2)-xy-not the same soo nope
origin-(-x^2)+xy=x- yyessss the graph will be symmetric
y=x-y^2+xy-nooot even close
now we will talk about graphing these things
if they give you f(x) , they could tell you this
1.-f(x)
2.f(-x)
3.lf(x)l
and i can't remember the last one if there is a last one
but if they tell you number 1, you would reflect the graph on the x axis
if they tell you 2. then you reflect on the y axis
if they tell you 3, then you make all negative y values positive and graph.
havee fuuuuuuuunnnnnnn withhh that.
These are the rules to see if something is symmetric
( i left my math binder at school so forgive me if i'm wrong)
to see if its symmetric on the x axis, you plug in negative y-and see if this equation and the original match
to see if its symmetric on the y axis, you plug in negative x-ditto
to see if its symmetric on the origin, you plug in negative x and y-ditto
to see if its symmetric ont the y=x, you switch the x and ys.-and ditto
example:
f(x)= x^2+xy
x axis-x^2-xy-not the same equation so noooooo
y axis-(-X^2)-xy-not the same soo nope
origin-(-x^2)+xy=x- yyessss the graph will be symmetric
y=x-y^2+xy-nooot even close
now we will talk about graphing these things
if they give you f(x) , they could tell you this
1.-f(x)
2.f(-x)
3.lf(x)l
and i can't remember the last one if there is a last one
but if they tell you number 1, you would reflect the graph on the x axis
if they tell you 2. then you reflect on the y axis
if they tell you 3, then you make all negative y values positive and graph.
havee fuuuuuuuunnnnnnn withhh that.
Nicala's Blog
This week in class we did the last section in chapter four and then we reviewed for the test on friday that i wasnt here to take. I dont remember to much about what we learned this week especially since we had a four day weekend and i didnt come friday but i do remember what i learned in physics with the triangles tht we also learned in physics so i will review that. In chapter nine section we learned how to find the legs, hyp and the angles in the right triangles using pythagorem thereom and SOHCAHTOA.
SOH-sine=opp/hyp
CAH-cosine=adj/hyp
TOA-tangent=opp/adj
example problem
in a ABC triangle, a=12, b=5, c=? solve using the pythagorem theorem= (a)squared+(b)squared=(c)squared
(12)squared+(5)squared=(c)squared
144+25=(c)squared
169=(c)squared
take the square root of 169
c=13
Friday, January 21, 2011
Taylor's 7th Blog Review
REVIEW TIME!!!
This is on the last part of my notes from section 7.1.
Now lets learn how to get minutes and seconds from our equations.
ab.cd
.cd x 60 = y minutes y'
y x60 = z seconds z''
This is the formula that we will be using in the following problem.
Covert 18.488 degrees to minutes and seconds.
.488 x 60= 29.28 First we will put .488 into the equation because it is after the decimal point. We will then multiply it by 60 so that we can get 29.28 . Since we have another decimal point we must continue the equation.
.28 x 60= 16.8 We will take the numbers after the decimal and multiply them by 60. Since we don't need the .8 you will simply drop it, since we just need seconds.
18 degrees 29' 16'' This is what your answer should be and what it should look like.
The final thing that you must know is how to convert the minutes and seconds back into degrees. You will use the following formula in order to do this.
x'=minutes y''= seconds
x/60 + y/3600= degrees
Now use it to solve the following problem.
Convert 28 degrees 54' 22'' in to degrees.
54/60 + 22/3600 First place the seconds and minutes in their appropriate location in the formula.
.906 Then you add the fractions together and convert the fraction that you may have gotten if you are doing this by hand into a decimal.
28.906 degrees Than you simply place the decimal behind the number and that is how you get your answer.
That is all the notes from the last part of section 7.1. Until my next blog BYE.
This is on the last part of my notes from section 7.1.
Now lets learn how to get minutes and seconds from our equations.
ab.cd
.cd x 60 = y minutes y'
y x60 = z seconds z''
This is the formula that we will be using in the following problem.
Covert 18.488 degrees to minutes and seconds.
.488 x 60= 29.28 First we will put .488 into the equation because it is after the decimal point. We will then multiply it by 60 so that we can get 29.28 . Since we have another decimal point we must continue the equation.
.28 x 60= 16.8 We will take the numbers after the decimal and multiply them by 60. Since we don't need the .8 you will simply drop it, since we just need seconds.
18 degrees 29' 16'' This is what your answer should be and what it should look like.
The final thing that you must know is how to convert the minutes and seconds back into degrees. You will use the following formula in order to do this.
x'=minutes y''= seconds
x/60 + y/3600= degrees
Now use it to solve the following problem.
Convert 28 degrees 54' 22'' in to degrees.
54/60 + 22/3600 First place the seconds and minutes in their appropriate location in the formula.
.906 Then you add the fractions together and convert the fraction that you may have gotten if you are doing this by hand into a decimal.
28.906 degrees Than you simply place the decimal behind the number and that is how you get your answer.
That is all the notes from the last part of section 7.1. Until my next blog BYE.
Charlie.
This week we kinda went over the whole chapter.
i think it's chapter 4.. that's not important
anyways, in the chapter we learned domain, range, inverse, zeros, and graphs.
*for polynomials..
the domain is always (- infinity, infinity)*
the range is only found if the exponent is odd, if it's odd the answer is (- infinity, infinity)*
to find zeros you set the whole problem equal to zero*
[examples] for domain, zero, and range of a polynomial:
1. y = x^2 + 5x + 6
Domain->
(- infinity, infinity)
Range ->
can't be found
Zeros ->
x^2 + 5x + 6 = 0
(x + 3) (x+2)
x = -3 & -2
2. y = 2x + 3
Domain ->
( - infinity, infinity)
Range ->
( - infinity, infinity)
Zeros ->
2x + 3 = 0
2x = -3
x = -3/2
3. y = x^2 + 2x -3
Domain ->
( - infinity, infinity)
Range ->
Can't be found
Zeros ->
x^2 + 2x - 3 = 0
(x + 3) (x-1)
x = -3 & 1
i think it's chapter 4.. that's not important
anyways, in the chapter we learned domain, range, inverse, zeros, and graphs.
*for polynomials..
the domain is always (- infinity, infinity)*
the range is only found if the exponent is odd, if it's odd the answer is (- infinity, infinity)*
to find zeros you set the whole problem equal to zero*
[examples] for domain, zero, and range of a polynomial:
1. y = x^2 + 5x + 6
Domain->
(- infinity, infinity)
Range ->
can't be found
Zeros ->
x^2 + 5x + 6 = 0
(x + 3) (x+2)
x = -3 & -2
2. y = 2x + 3
Domain ->
( - infinity, infinity)
Range ->
( - infinity, infinity)
Zeros ->
2x + 3 = 0
2x = -3
x = -3/2
3. y = x^2 + 2x -3
Domain ->
( - infinity, infinity)
Range ->
Can't be found
Zeros ->
x^2 + 2x - 3 = 0
(x + 3) (x-1)
x = -3 & 1
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?
Sunday, January 16, 2011
Nicala's Post
In class we are in chapter four. In chapter four section two we are doing operation on functions also known as notation. There are four different different types of function: sum of f and g: (f+g)(x)=f(x)+g(x)
difference of and g:(f -g)(x)=f(x)-g(x)
product of f ad g: (fxg)(x)=f(x
)xg(x)
Examples
f(x)=4x+1 g(x)=3-x
(f+g)(x)= (4x+1)+(3-x)= 3x+4
TORIS.
WOOOOOO MATHH!
1. Polynomials:
Domain: (-∞, ∞)
Range: (-∞, ∞)
**This only applies to polynomials whose highest exponent is odd.
2. Square Roots
There are three steps to finding the domain.
1. Set the inside equal to 0 and solve for x
2. Set up a number line
3. Plug in the numbers and use the non-negative intervals
Range: [vertical shift, infinity)
3. Fractions
Two steps for domain and range.
Domain:
1. Set bottom equal to zero and solve for x
2. Set up intervals
Range:
1. Take limit as x > ∞
2. Set up intervals
4. Absolute value
Domain: (-∞, ∞)
Range: [shift, ∞) or (-∞, shift]
Malorie's blog
4-3 Symmetry
x-axis: 1. plug in (-y)
2.simplify
3.if equations are = then it has symmetry.
y-axis: 1. plug in (-x)
2. same as first
3. same as first
origin: 1. plug in (-x) and (-y)
2. same as first
3. same as first
y=x: 1. switch x and y
2. solve for y
3. same as first
Ex:
y^2+xy=10, for symmetry about the x-axis, y-axis, origin, and y=x.
x-axis: (-y)^2+x(-y)=10
y^2-xy=10
nooooooo!
y-axis: y^2+(-x)y=10
y^2-xy=10
noooopee!
origin: (-y)^2+(-x)(-y)=10
y^2+xy=10
YEEEEES (:
y=x: x^2+xy=10
Noooo shot
so the final answer would be that only the origin has symmetry to the beginning equation
x-axis: 1. plug in (-y)
2.simplify
3.if equations are = then it has symmetry.
y-axis: 1. plug in (-x)
2. same as first
3. same as first
origin: 1. plug in (-x) and (-y)
2. same as first
3. same as first
y=x: 1. switch x and y
2. solve for y
3. same as first
Ex:
y^2+xy=10, for symmetry about the x-axis, y-axis, origin, and y=x.
x-axis: (-y)^2+x(-y)=10
y^2-xy=10
nooooooo!
y-axis: y^2+(-x)y=10
y^2-xy=10
noooopee!
origin: (-y)^2+(-x)(-y)=10
y^2+xy=10
YEEEEES (:
y=x: x^2+xy=10
Noooo shot
so the final answer would be that only the origin has symmetry to the beginning equation
Nathan's blog
So, we went back to algebra 2 for the new year and the concepts are very easy, but as always with B-Rob, every thing gets pushed to the next level and it can get difficult at some points. In section 1, we learned about domain and range, section 2 was on.......i forgot, and section 3 dealt with symmetry and all that good stuff.
There were some formulas that you need to know for section 3:
1.) Y-axis - You have to plug in (-x)
2.) X-axis - You have to plug in (-y), then simplifiy; if the equations are equal then it has symmetry.
3.) Origin - You have to plug in (-x) and (-y)
4.) Y=X - You have to switch x and y, and then solve for y.
Section 4-1 dealt with domain and range, and finding the zeros.
y=4x^2+6x-7
D:(-∞,∞)
R:
Zeros:(x+7)(x-1)
x=-7,1
That's all for this blog, see everyone on Tuesday.
There were some formulas that you need to know for section 3:
1.) Y-axis - You have to plug in (-x)
2.) X-axis - You have to plug in (-y), then simplifiy; if the equations are equal then it has symmetry.
3.) Origin - You have to plug in (-x) and (-y)
4.) Y=X - You have to switch x and y, and then solve for y.
Section 4-1 dealt with domain and range, and finding the zeros.
y=4x^2+6x-7
D:(-∞,∞)
R:
Zeros:(x+7)(x-1)
x=-7,1
That's all for this blog, see everyone on Tuesday.
Lawrence's blog
for the blog this week i will go over 4-3. this section has to do with symmetry. it is fairly easy but you have to know the formulas like always. im getting really tired of having to remember this crap but hey it helps me pass some of the quizes and tests so i cant really complain. they have four formulas to use and it deals with drawing so i wont be able to draw them on a computer obviously so i will just give you the formulas and if you need help i can show you how to do these.
Y-AXIS 1: plug in (-x)
2:
3:
X-AXIS 1: plug in (-y)
2: simplify
3: if equations are equal then it has symmetry
ORIGIN 1: plug in (-x) and (-y)
2:'
3:'
Y=X 1: switch x and y
2: solve for y
3: '
Y-AXIS 1: plug in (-x)
2:
3:
X-AXIS 1: plug in (-y)
2: simplify
3: if equations are equal then it has symmetry
ORIGIN 1: plug in (-x) and (-y)
2:'
3:'
Y=X 1: switch x and y
2: solve for y
3: '
Friday, January 14, 2011
TAYLOR'S BLOG REVIEW
REVIEW TIME!!!
THIS IS A REVIEW ON PART 2 OF MY SECTION 7.1 NOTES!!
There are some angles that are called coterminal angles. Coterminal means that the degrees 'spins' around the angle in this usage.
If you are asked to find the positive or/and negative form of a coterminal angle for degrees you will use this formula.
degrees +/- 360 degrees
Yet, if you are asked to find the positive or/and negative form of a coterminal angle for radians you would use this formula.
rads +/- 2 PI
An example of these situations are as followed.
Find a positive coterminal angle of 13 degrees.
13 degrees + 360 degrees First, you will place 13 degrees into the equation and since we need it to be positive we will add 360 to it.
373 degrees This is the correct answer that you should have gotten after adding 360 to 13.
Find the negative coterninal angle of 32 degrees.
32 degrees - 360 degrees First place 32 into its correct place in the equation and since we need to find the negative coterminal angle we will subtract 360 from 32 instead of adding it.
-328 degrees This is the negative answer that you should have gotten.
Now lets find the positive coterminal angle for 8PI/10.
8PI/10 + 2PI First place the 8PI/10 in its correct location in the equation and since we need to find the positive coterminal angle you will add 2Pi.
8PI/10 + 20pi/10 Now convert the 2PI so that you can add correctly. Do this by multiplying 2 by 10 and placing a ten in the denominator.
28PI/10 After you do that you add the fractions together.
14PI/5 Then you reduce the fraction by dividing it by two.
Now lets find the negative coterminal angle of 2PI/8.
2PI/8 - 2PI First place 2Pi/8 in its proper location in the equation and since we are finding the negative coterminal angle you will subtract 2PI instead of adding it.
2PI/8 - 16PI/8 Than you will make 2PI into a fraction by placing it over 8 and multiplying it by 8.
-14PI/8 You will then subtract and get this answer.
-7PI/4 Then you will reduce the fraction by diving it by two and this is your answer.
This is the second part from my 7.1 notes.
THIS IS A REVIEW ON PART 2 OF MY SECTION 7.1 NOTES!!
There are some angles that are called coterminal angles. Coterminal means that the degrees 'spins' around the angle in this usage.
If you are asked to find the positive or/and negative form of a coterminal angle for degrees you will use this formula.
degrees +/- 360 degrees
Yet, if you are asked to find the positive or/and negative form of a coterminal angle for radians you would use this formula.
rads +/- 2 PI
An example of these situations are as followed.
Find a positive coterminal angle of 13 degrees.
13 degrees + 360 degrees First, you will place 13 degrees into the equation and since we need it to be positive we will add 360 to it.
373 degrees This is the correct answer that you should have gotten after adding 360 to 13.
Find the negative coterninal angle of 32 degrees.
32 degrees - 360 degrees First place 32 into its correct place in the equation and since we need to find the negative coterminal angle we will subtract 360 from 32 instead of adding it.
-328 degrees This is the negative answer that you should have gotten.
Now lets find the positive coterminal angle for 8PI/10.
8PI/10 + 2PI First place the 8PI/10 in its correct location in the equation and since we need to find the positive coterminal angle you will add 2Pi.
8PI/10 + 20pi/10 Now convert the 2PI so that you can add correctly. Do this by multiplying 2 by 10 and placing a ten in the denominator.
28PI/10 After you do that you add the fractions together.
14PI/5 Then you reduce the fraction by dividing it by two.
Now lets find the negative coterminal angle of 2PI/8.
2PI/8 - 2PI First place 2Pi/8 in its proper location in the equation and since we are finding the negative coterminal angle you will subtract 2PI instead of adding it.
2PI/8 - 16PI/8 Than you will make 2PI into a fraction by placing it over 8 and multiplying it by 8.
-14PI/8 You will then subtract and get this answer.
-7PI/4 Then you will reduce the fraction by diving it by two and this is your answer.
This is the second part from my 7.1 notes.
Sunday, January 9, 2011
Feroz's Blog
Ehhhhh. So tired. Went to that Math lecture with Mary and all them. It was fairly interesting. Anyway, I'm gonna cover some stuff from Chapter 4. So yeah.
I copied these down in my notebook. They look important. I dunno.
1. Polynomials:
Domain: (-∞, ∞)
Range: (-∞, ∞)*
*This only applies to polynomials whose highest exponent is odd.
2. Square Roots
There are three steps to finding the domain.
1. Set the inside equal to 0 and solve for x
2. Set up a number line
3. Plug in the numbers and use the non-negative intervals
Range: [vertical shift, infinity)
3. Fractions
Two steps for domain and range.
Domain:
1. Set bottom equal to zero and solve for x
2. Set up intervals
Range:
1. Take limit as x > ∞
2. Set up intervals
4. Absolute value
Domain: (-∞, ∞)
Range: [shift, ∞) or (-∞, shift]
That was fun.
I copied these down in my notebook. They look important. I dunno.
1. Polynomials:
Domain: (-∞, ∞)
Range: (-∞, ∞)*
*This only applies to polynomials whose highest exponent is odd.
2. Square Roots
There are three steps to finding the domain.
1. Set the inside equal to 0 and solve for x
2. Set up a number line
3. Plug in the numbers and use the non-negative intervals
Range: [vertical shift, infinity)
3. Fractions
Two steps for domain and range.
Domain:
1. Set bottom equal to zero and solve for x
2. Set up intervals
Range:
1. Take limit as x > ∞
2. Set up intervals
4. Absolute value
Domain: (-∞, ∞)
Range: [shift, ∞) or (-∞, shift]
That was fun.
kaitlyn's bloggggg
In this blogggg, i will go overr Chapter 14-1. This section is all about domain and range. I've learned this before in algebra 2 i thinkk, but i could never get it and thought it was the hardest thing to do, but now i understand it pretty well now that it was broken down more for me.
There are different ways to find the domain and range with different equations:
Polynomials- domain: always (-infinity,infinity)
range: (-infinity,infinity) if odddd
Square roots- domain: 1)set inside equal to 0 and solve for x
2)set up a number line
3)plug in the numbers, use the non-negative intervals
range: [vertical shift, infinity)
Fractions- domain: 1)set bottom equal to zero and solve for x
2) set up intervals
range: 1)take limit as x->infinity
2)set up intervals
Absolute value- domain: (-infinity, infinity)
range: [shift, infinity) or (-infinity, shift]
Example: x^3+2x-3
domain: (-infinity, infinity)
range: (-infinity, infinity)
Example: (squarerootof x-3)+1
x-3=0
x=0
domain: (-infinity, 3) u (3, infinity)
range: [1, infinity)
There are different ways to find the domain and range with different equations:
Polynomials- domain: always (-infinity,infinity)
range: (-infinity,infinity) if odddd
Square roots- domain: 1)set inside equal to 0 and solve for x
2)set up a number line
3)plug in the numbers, use the non-negative intervals
range: [vertical shift, infinity)
Fractions- domain: 1)set bottom equal to zero and solve for x
2) set up intervals
range: 1)take limit as x->infinity
2)set up intervals
Absolute value- domain: (-infinity, infinity)
range: [shift, infinity) or (-infinity, shift]
Example: x^3+2x-3
domain: (-infinity, infinity)
range: (-infinity, infinity)
Example: (squarerootof x-3)+1
x-3=0
x=0
domain: (-infinity, 3) u (3, infinity)
range: [1, infinity)
woooo another bloogg! yahhhh. SIKKEE!
so for the first lesson of the year we went back to algebra 2, too bad i dont remember learning this in algebra 2 so i was lost as a buried treasure during this week. i tried to do my homework but all the problems just didnt make sense.. ughh, sorry BRob, but Advanced Math STINKS!
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 the non-negative intervals.
Range: (vertical shift, infinity)
*If √#-x^2 Range=(0,√#)
3. Fractions
Domain:
1.) Set bottom equal to 0, and solve for x.
2.) Set up intervals.
Range:
1.) Take limit as x > ∞2.) Set up intervals.
Ex. 3/x-2
x-2=0
x=2
D: (-∞,2) u (2,∞)
R: lim/x>∞ 3/x-2=0
R: (-∞,0) u (0,∞)
THAT IS ALL.
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 the non-negative intervals.
Range: (vertical shift, infinity)
*If √#-x^2 Range=(0,√#)
3. Fractions
Domain:
1.) Set bottom equal to 0, and solve for x.
2.) Set up intervals.
Range:
1.) Take limit as x > ∞2.) Set up intervals.
Ex. 3/x-2
x-2=0
x=2
D: (-∞,2) u (2,∞)
R: lim/x>∞ 3/x-2=0
R: (-∞,0) u (0,∞)
THAT IS ALL.
Helen's Blog
Finding Domain and Range:
1) Polynomial- an equation with no variables in the denominator.
domain: always (- infinity, infinity)
range: (-infinity, infinity) if the largest exponent is odd
2) Square Roots
domain: a) set inside = 0 and solve for x
b) set up a number line
c) plug in the numbers, use the non-negative interval
range: [vertical shift, infinity)
3) Fractions
domain: a) set bottom = 0 solve for x
b) set up intervals
range: a) take limit as x-> infinity
b) set up intervals
4) Absolute Value
domain: ( -infinity, infinity)
range: [ shift, infinity) or ( -infinity, shift]
Examples:
1) 3x^3 + 4x^2-7x
domain: ( - infinity, infinity)
range: ( - inifinity, infinity)
2) x+1/ x^2+5x+6
D: x^2+5x+6
(x+3)(x+2)
D: (- infinity, -3)u(-3,-2)u(-2, infinity)
R: ( -infinity,0)u(0,infinity)
1) Polynomial- an equation with no variables in the denominator.
domain: always (- infinity, infinity)
range: (-infinity, infinity) if the largest exponent is odd
2) Square Roots
domain: a) set inside = 0 and solve for x
b) set up a number line
c) plug in the numbers, use the non-negative interval
range: [vertical shift, infinity)
3) Fractions
domain: a) set bottom = 0 solve for x
b) set up intervals
range: a) take limit as x-> infinity
b) set up intervals
4) Absolute Value
domain: ( -infinity, infinity)
range: [ shift, infinity) or ( -infinity, shift]
Examples:
1) 3x^3 + 4x^2-7x
domain: ( - infinity, infinity)
range: ( - inifinity, infinity)
2) x+1/ x^2+5x+6
D: x^2+5x+6
(x+3)(x+2)
D: (- infinity, -3)u(-3,-2)u(-2, infinity)
R: ( -infinity,0)u(0,infinity)
Nathan's blog
For the start of the new year, we went back to Algebra 2 with domain, range, and all that other good stuff. So, I will review chapter 4 section 1.
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 the non-negative intervals.
Range: (vertical shift, infinity)
*If √#-x^2 Range=(0,√#)
3. Fractions
Domain: 1.) Set bottom equal to 0, and solve for x.
2.) Set up intervals.
Range: 1.) Take limit as x > ∞
2.) Set up intervals.
Ex. 3/x-2
x-2=0
x=2
D: (-∞,2) u (2,∞)
R: lim/x>∞ 3/x-2=0
R: (-∞,0) u (0,∞)
That's all for this blog. I guess this helped.
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 the non-negative intervals.
Range: (vertical shift, infinity)
*If √#-x^2 Range=(0,√#)
3. Fractions
Domain: 1.) Set bottom equal to 0, and solve for x.
2.) Set up intervals.
Range: 1.) Take limit as x > ∞
2.) Set up intervals.
Ex. 3/x-2
x-2=0
x=2
D: (-∞,2) u (2,∞)
R: lim/x>∞ 3/x-2=0
R: (-∞,0) u (0,∞)
That's all for this blog. I guess this helped.
Blog number 1 of the new yearr!
well, this week, we learned about domain and range of different equations. We are learning this and we learned how to do all of them EXCEPT the one we had on the team test! how lucky are we?!? but that doesn't matter, we won anyway, even if we were against ourselves. woohoo. hahahahhaa. Honestly, my whole highschool experience has revolved around math. Every other class seems like a joke. I don't feel like i'm learning anything else, and i'm spending my weekends learning more math. It's really interesting, that thing I just went to on sunday was realllyyy coool, and i wish we could have a follow up because it was interesting. Just saying, i don't like all the homework and everything but in a way i like it all. ANYWAY, this week, domain and range.
donmain range for a polynomial
always negative infinity, infinity
range WE DON'T KNOWWW!!!
domain and range for fractions.
oh crap, gotta bust out the notes now.
domain-set bottom equal to zero and solve for x(then intervals)
range-take the limit(then intervals)
square roots
domain-set inside=0 and solve for x
set up a number line
seee which side works
absolute value
domain-negative infinity to infinity
range, shift , infinity, or neg infinity, shift.
the easiets one is polynomial so here's an example.
5x^5+6x+1
domain is alwayys neg infinity, infinity
fraction:
3/x-2
x-2=0
x=2
(neg infinity, 2,) u (2, infinity)
range:the bottom is bigger so limit it iss 0
(neg infinity, 0) u ( o, infinity)
and there's two examples
this week was KINDA easy.
it was whack when we learned this and the homework didn't correspond.
but ha.
bye.
donmain range for a polynomial
always negative infinity, infinity
range WE DON'T KNOWWW!!!
domain and range for fractions.
oh crap, gotta bust out the notes now.
domain-set bottom equal to zero and solve for x(then intervals)
range-take the limit(then intervals)
square roots
domain-set inside=0 and solve for x
set up a number line
seee which side works
absolute value
domain-negative infinity to infinity
range, shift , infinity, or neg infinity, shift.
the easiets one is polynomial so here's an example.
5x^5+6x+1
domain is alwayys neg infinity, infinity
fraction:
3/x-2
x-2=0
x=2
(neg infinity, 2,) u (2, infinity)
range:the bottom is bigger so limit it iss 0
(neg infinity, 0) u ( o, infinity)
and there's two examples
this week was KINDA easy.
it was whack when we learned this and the homework didn't correspond.
but ha.
bye.
Lawrence's blog
We went all the way back to chapter 4. it is a review from algebra 2 last year. i found section 1 to be kinda easy but im still working on trying to get better at. section 1 deals with finding the domain and range. there are four ways to do this.
1. Polynomial- they have no variable in denominator
the domain will always be (-infinity, infinity)
the range will always be (-infinity, infinity) if the degree of the polynomial is odd otherwise you will not have a range.
2. Square roots-
domain-you have to set equal to zero and solve for x.
then you put it on a number line
and plug in the numbers
range- use a [vertical shift, infinity)
3. Fractions
Domain- set the bottom equal to zero and solve for x
then set up intervals
Range- take limits x(infinity)
4. Absolute value
Domain: (-infinity, infinity)
Range: ( zero, infinity)
example:3x3+4x2-7x
domain: ( -infinity, infinity)
range: (-infinity, infinity)
i hope this helped some of yall better understand this chapter and section. if you need help just ask me. i get this chapter for the most part.
1. Polynomial- they have no variable in denominator
the domain will always be (-infinity, infinity)
the range will always be (-infinity, infinity) if the degree of the polynomial is odd otherwise you will not have a range.
2. Square roots-
domain-you have to set equal to zero and solve for x.
then you put it on a number line
and plug in the numbers
range- use a [vertical shift, infinity)
3. Fractions
Domain- set the bottom equal to zero and solve for x
then set up intervals
Range- take limits x(infinity)
4. Absolute value
Domain: (-infinity, infinity)
Range: ( zero, infinity)
example:3x3+4x2-7x
domain: ( -infinity, infinity)
range: (-infinity, infinity)
i hope this helped some of yall better understand this chapter and section. if you need help just ask me. i get this chapter for the most part.
Friday, January 7, 2011
Nicala's Blog
this week in advanced we went back a couple of chapters to chapter 4 section 1. we learned the domain, range, and zeros which you should remember from algebra 2.
there are four different problem are
1. Polynomial- they have no variable in denominator
the domain will always be (-infinity, infinity)
the range will always be (-infinity, infinity) if the degree of the polynomial is odd otherwise you will not have a range.
2. Square roots-
domain-you have to set equal to zero and solve for x.
then you put it on a number line
and plug in the numbers
range- use a [vertical shift, infinity)
3. Fractions
Domain- set the bottom equal to zero and solve for x
then set up intervals
Range- take limits x(infinity)
4. Absolute value
Domain: (-infinity, infinity)
Range: ( zero, infinity)
Example Problem
3x^3+5x^2+9x
domain:( -infinity, infinity)
and because the degree of the polynomial is 3 the range is (-infinity, infinity)
there are four different problem are
1. Polynomial- they have no variable in denominator
the domain will always be (-infinity, infinity)
the range will always be (-infinity, infinity) if the degree of the polynomial is odd otherwise you will not have a range.
2. Square roots-
domain-you have to set equal to zero and solve for x.
then you put it on a number line
and plug in the numbers
range- use a [vertical shift, infinity)
3. Fractions
Domain- set the bottom equal to zero and solve for x
then set up intervals
Range- take limits x(infinity)
4. Absolute value
Domain: (-infinity, infinity)
Range: ( zero, infinity)
Example Problem
3x^3+5x^2+9x
domain:( -infinity, infinity)
and because the degree of the polynomial is 3 the range is (-infinity, infinity)
Taylor 4th Blog Review
REVIEW TIME!!!!
In the following lesson I shall be informing you on part of lesson 7.1 of my Advanced Math class.
There are two units of measure for angles. They are degrees and radians.
If you want to convert degrees to radians you must use the following formula.
degrees x PI/180
An example of this would be the following problem.
Convert 25 degrees to radians.
25 degrees x PI/180 1st. Place the degrees ,25, into its proper location in the equation.
25PI/180 You will then multiply 25 degrees and PI/180, or 25 X PI to get this answer.
5PI/36 You will then divide 180 into 25,or reduce, and put it in a fraction form. This is your answer.
IMPORTANT NOTE: PI IS LIKE X A VARIABLE SO DON'T PUT IT INTO YOUR CALCULATOR OR YOU WILL GET THE WRONG ANSWER!
If you want to convert radians to degrees you would use this formula.
rads x 180/PI
An example of this equation has been given.
Convert 20PI to degrees.
20PI X 180/PI First place your radians into its proper place in the equation.
20P/I X `180/P/I Since 20PI and 180/PI both have PI and one is in the lower part of the fraction and the other is in the upper part of the fraction. We can cancel out PI in the equation.
20 X 180 You will then multiply 20 times 180.
3600 degrees This is what your answer should be.
IMPORTANT NOTE: MAKE SURE TO PUT THE DEGREES MARK OR IT WILL BE WRONG!!
That is all the equations, formula and information that we learned in the first part of section 7.1. Until my next blog BYE.
In the following lesson I shall be informing you on part of lesson 7.1 of my Advanced Math class.
There are two units of measure for angles. They are degrees and radians.
If you want to convert degrees to radians you must use the following formula.
degrees x PI/180
An example of this would be the following problem.
Convert 25 degrees to radians.
25 degrees x PI/180 1st. Place the degrees ,25, into its proper location in the equation.
25PI/180 You will then multiply 25 degrees and PI/180, or 25 X PI to get this answer.
5PI/36 You will then divide 180 into 25,or reduce, and put it in a fraction form. This is your answer.
IMPORTANT NOTE: PI IS LIKE X A VARIABLE SO DON'T PUT IT INTO YOUR CALCULATOR OR YOU WILL GET THE WRONG ANSWER!
If you want to convert radians to degrees you would use this formula.
rads x 180/PI
An example of this equation has been given.
Convert 20PI to degrees.
20PI X 180/PI First place your radians into its proper place in the equation.
20P/I X `180/P/I Since 20PI and 180/PI both have PI and one is in the lower part of the fraction and the other is in the upper part of the fraction. We can cancel out PI in the equation.
20 X 180 You will then multiply 20 times 180.
3600 degrees This is what your answer should be.
IMPORTANT NOTE: MAKE SURE TO PUT THE DEGREES MARK OR IT WILL BE WRONG!!
That is all the equations, formula and information that we learned in the first part of section 7.1. Until my next blog BYE.
Charlie's Weekend Blog.
in advanced math this week we started looking for zeros, functions, domain, and range. we had learned this back in the gap in algebra two.. i know what i'm doing, (for the most part).
there are 3 situations: parabolas, square roots, & fractions.
**in this blog we're just doing parabolas**
to find the domain of the parabolas, the answer is always (-infinity, infinity).
to find the range for the parabolas, the answer is (-infinity, infinity). but only if the largest exponent is odd.
to find the zero, you set the whole given problem up equal to zero.
then, to find if it is a function you plug it into the calculater. & sketch the graph on the paper. then do the straight line test. if it doesn't pass through the graphed shape more than once, then it is a function.
example:
f(x) = 6x + 2
Zeros > -1/3
domain > (-infinity, infinity)
Range > (-infinity, infinity)
function? > yes.
6x + 2 = 0
x = -2/6 = -1/3
there are 3 situations: parabolas, square roots, & fractions.
**in this blog we're just doing parabolas**
to find the domain of the parabolas, the answer is always (-infinity, infinity).
to find the range for the parabolas, the answer is (-infinity, infinity). but only if the largest exponent is odd.
to find the zero, you set the whole given problem up equal to zero.
then, to find if it is a function you plug it into the calculater. & sketch the graph on the paper. then do the straight line test. if it doesn't pass through the graphed shape more than once, then it is a function.
example:
f(x) = 6x + 2
Zeros > -1/3
domain > (-infinity, infinity)
Range > (-infinity, infinity)
function? > yes.
6x + 2 = 0
x = -2/6 = -1/3
Sunday, January 2, 2011
Feroz's Final Holiday Blog
Last one. Go me.
Chapter 13 -1
Sequences:
Arithmetic = A sequence where you're adding the same number in a pattern.
Geometric = Same thing just with multiplying.
ex. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21....
That would be Arithmetic.
ex. 1, 3, 9, 27, 81....
That would be Geometric.
Anyway, formulas for finding the Nth term:
Arith. = tn= t1+ (n-1)d
Geo. = tn = t1 x r^n-1
ex. 7, 10, 13, 16...
It's arithmetic, so use the arith. formula.
7 + 3n - 3
= 3n + 4
Finally done with this. Time to get some sleep.
Chapter 13 -1
Sequences:
Arithmetic = A sequence where you're adding the same number in a pattern.
Geometric = Same thing just with multiplying.
ex. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21....
That would be Arithmetic.
ex. 1, 3, 9, 27, 81....
That would be Geometric.
Anyway, formulas for finding the Nth term:
Arith. = tn= t1+ (n-1)d
Geo. = tn = t1 x r^n-1
ex. 7, 10, 13, 16...
It's arithmetic, so use the arith. formula.
7 + 3n - 3
= 3n + 4
Finally done with this. Time to get some sleep.
Feroz's Holiday Blog #3
SOHCAHTOA.
This has to do with triangles. And trig functions. Yeah.
Sin = Opposite/Hypotenuse, y/r
Cos = Adjacent/Hypotenuse, x/r
Tan = Opposite/Adjacent, y/x
The other three are just the opposite of those three.
Csc = Hypotenuse/Opposite, r/y
Sec = Hypotenuse/Adjacent, r/x
Cot = Adjacent/Opposite, x/y
This comes in handy a lot, and it's not hard to remember.
Oh, and I forgot to mention this only works for right triangles.
This has to do with triangles. And trig functions. Yeah.
Sin = Opposite/Hypotenuse, y/r
Cos = Adjacent/Hypotenuse, x/r
Tan = Opposite/Adjacent, y/x
The other three are just the opposite of those three.
Csc = Hypotenuse/Opposite, r/y
Sec = Hypotenuse/Adjacent, r/x
Cot = Adjacent/Opposite, x/y
This comes in handy a lot, and it's not hard to remember.
Oh, and I forgot to mention this only works for right triangles.
Feroz's Holiday Blog #2
blah blah blah Chapter 10-1 blah blah blah
Formulas:
sin: (a +/- b) = sinacosb +/- cosasinb
cos: (a +/- b) = cosacosb +/- sinasinb
ex. sin 75
= (sin 45 + sin 35) = sin45cos35 + cos45sin35
= (√2/2) (√3/2) + (√2/2)(1/2)
= √6 + √2
------------
4
Formulas:
sin: (a +/- b) = sinacosb +/- cosasinb
cos: (a +/- b) = cosacosb +/- sinasinb
ex. sin 75
= (sin 45 + sin 35) = sin45cos35 + cos45sin35
= (√2/2) (√3/2) + (√2/2)(1/2)
= √6 + √2
------------
4
Dom's blog # 4
chapter 13 -1.
formulas:
Arithmetic = tn = t1 (n-1)d
Geometric = tn = t1 x r^n-1
ex. find first 3 terms for tn = 201n
t1 = 201(1) 3
201 3 = 204
t2 = 201(2) 3
402 3 = 405
t3 = 201(3) 3
603 3 = 606
ex. find formula for n^m term :3,7,11,15 (Arithmetic)
tn = t1 (n-1)d
tn = 3 (n-1)(4)
tn = 3 4n - 4
tn = 4n-1
formulas:
Arithmetic = tn = t1 (n-1)d
Geometric = tn = t1 x r^n-1
ex. find first 3 terms for tn = 201n
t1 = 201(1) 3
201 3 = 204
t2 = 201(2) 3
402 3 = 405
t3 = 201(3) 3
603 3 = 606
ex. find formula for n^m term :3,7,11,15 (Arithmetic)
tn = t1 (n-1)d
tn = 3 (n-1)(4)
tn = 3 4n - 4
tn = 4n-1
Dom's blog # 3
chapter 11- 1
converting to rectangular
x=r cos theta
y=r sin theta
and convert to polar use the formulas
r=squared root x squared plus y squared
Example: Give the polar point for (3,5)
r=squared root 3 squared plus 5 squared= squared root of 34
tan theta=three over 5
theta= tan inverse=31
theta=211
= squared root of 34, 31 and the negative squared root of 34,211
converting to rectangular
x=r cos theta
y=r sin theta
and convert to polar use the formulas
r=squared root x squared plus y squared
Example: Give the polar point for (3,5)
r=squared root 3 squared plus 5 squared= squared root of 34
tan theta=three over 5
theta= tan inverse=31
theta=211
= squared root of 34, 31 and the negative squared root of 34,211
Dom's blog # 1
Chapter 8-5
Divide by a trig function when solving to cancel.
Cancel from the inside of a trig function.
Ex: sin(2x)/2 = you CAN'T cancel the 2
Ex:
1)2sin^2 theta-1=0
2sin^2 theta=1
sin^2 theta=1/2
sin theta = /- sq.root of 1/2
theta= sin inverse (sq. root of 1/2)
2)sin^2x-sinx=cos^2 x
sin^2x-sinx=1-sin^2x
2sin^2x-sinx-1=0
(2sin^2 x -2sinx) (sinx-1)=0
2sinx(sinx-1) (sinx-1)=0
(2sinx 1)(sinx-1)=0
2sinx 1=0 sinx-1=0
sinx=-1/2 sinx=1
x=sin inverse (1/2) x=sin inverse (1)
x=7Ï€/6,11Ï€/6,Ï€/2 x=Ï€/2
3)sinxtanx=3sinx
sinxtanx-3sinx=0
sinx(tanx-3)=0
sinx=0 tanx-3=0
x=sin inverse(0) x=tan inverse (3)
Divide by a trig function when solving to cancel.
Cancel from the inside of a trig function.
Ex: sin(2x)/2 = you CAN'T cancel the 2
Ex:
1)2sin^2 theta-1=0
2sin^2 theta=1
sin^2 theta=1/2
sin theta = /- sq.root of 1/2
theta= sin inverse (sq. root of 1/2)
2)sin^2x-sinx=cos^2 x
sin^2x-sinx=1-sin^2x
2sin^2x-sinx-1=0
(2sin^2 x -2sinx) (sinx-1)=0
2sinx(sinx-1) (sinx-1)=0
(2sinx 1)(sinx-1)=0
2sinx 1=0 sinx-1=0
sinx=-1/2 sinx=1
x=sin inverse (1/2) x=sin inverse (1)
x=7Ï€/6,11Ï€/6,Ï€/2 x=Ï€/2
3)sinxtanx=3sinx
sinxtanx-3sinx=0
sinx(tanx-3)=0
sinx=0 tanx-3=0
x=sin inverse(0) x=tan inverse (3)
Dom's blog # 1
chapter 7-2
all you have to do is remember the formulas and the hints to pass this section
formulas
s=r(theta)
k-one half(r)squaredtheta
k=onehalf (r)(s)
s=arc length
r=radius
theta=central angle
k=area
theta=apparent size
s=diameter
r=distance between objects
Ex: A sector of a circle has an arc length of 8 cm and the area of 75 cm. Find its radius and the measure of its central angle.
s=8cm
k=75 cm
r=
theta=
using one of the formulas for area
k=one half(r)(s)
75=(1/2)(r)(8)
75=4r
then you divide four by both sides
r=18.75cm
next you use the arclength formula
s=(r)theta
8=(18.75)(theta)
theta=2.34375 rads
Feroz's Holiday Blog #1
This is Feroz doing his holiday blogs.
#1 Chapter 7
Simplest chapter in the book.
I guess I'll just go over converting to radians.
To radians: pi/180
To degrees: 180/pi
ex. 2pi
= 2 x 180 = 360 degrees
ex. 540 degrees
= 540/180 = 3pi
Ok, one lesson down.
#1 Chapter 7
Simplest chapter in the book.
I guess I'll just go over converting to radians.
To radians: pi/180
To degrees: 180/pi
ex. 2pi
= 2 x 180 = 360 degrees
ex. 540 degrees
= 540/180 = 3pi
Ok, one lesson down.
Tori Holiday Blog #4
DONEE!!!!!!!
the definition of polar is to graph using angles
polar form is (r, theta) and rectangular form is (x, y)
the formulas we used for this were...
r = `the square root of` x^2 + y^2 and
tantheta = (y/x) [ this easily changes to theta = tangent inverse of (y/x) ]
so to do this you basically just plug in the (x, y) rectangular form numbers into the two formulas to get the (r, theta) polar form.
EXAMPLE:
#1. turn rectangular form (-1, 2) into polar form
[ first, you plug the -1 and the 2 into the formulas ]
r = `the square root of` -1^2 + 2^2
= `the square root of` 1 + 4= +/- `the square root of` 5theta
= tangent inverse of (2/-1)= 63.565
[ tangent is negative on the unit circle in quadrant 2 &4, so now you make 63 negative and add 180 & make 63 negative and add 360 to get to those quadrants ]....
survey says:: `the square root of` 5, 116. 565.... `the square root of` -5, 296.565
the definition of polar is to graph using angles
polar form is (r, theta) and rectangular form is (x, y)
the formulas we used for this were...
r = `the square root of` x^2 + y^2 and
tantheta = (y/x) [ this easily changes to theta = tangent inverse of (y/x) ]
so to do this you basically just plug in the (x, y) rectangular form numbers into the two formulas to get the (r, theta) polar form.
EXAMPLE:
#1. turn rectangular form (-1, 2) into polar form
[ first, you plug the -1 and the 2 into the formulas ]
r = `the square root of` -1^2 + 2^2
= `the square root of` 1 + 4= +/- `the square root of` 5theta
= tangent inverse of (2/-1)= 63.565
[ tangent is negative on the unit circle in quadrant 2 &4, so now you make 63 negative and add 180 & make 63 negative and add 360 to get to those quadrants ]....
survey says:: `the square root of` 5, 116. 565.... `the square root of` -5, 296.565
Tori Holiday Blog #3
6 down, 1 to go!!
Section 10-1
Formulas:
cos(alpha +/- beta) cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta) sin alpha cos beta +/- cos alpha sin beta
Examples:
Ex1:Find the exact value of sin15
sin(45-30)=sin45cos30-cos45sin30=
(√2/2)(√3/2)-(√2/2)(1/2)=
√6-√2/4
Section 10-2 ( same concept but tan instead)
Formula:
tan(alpha + beta)= tan (alpha) + tan (beta) / 1-tan(alpha) tan(beta)
tan(alpha - beta)= tan(alpha)- tan (beta) / 1+tan(alpha)tan(beta)
Ex2:Simplify: tan27+tan18/1-tan27tan18=
tan(27+18)=tan45
=1
Section 10-1
Formulas:
cos(alpha +/- beta) cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta) sin alpha cos beta +/- cos alpha sin beta
Examples:
Ex1:Find the exact value of sin15
sin(45-30)=sin45cos30-cos45sin30=
(√2/2)(√3/2)-(√2/2)(1/2)=
√6-√2/4
Section 10-2 ( same concept but tan instead)
Formula:
tan(alpha + beta)= tan (alpha) + tan (beta) / 1-tan(alpha) tan(beta)
tan(alpha - beta)= tan(alpha)- tan (beta) / 1+tan(alpha)tan(beta)
Ex2:Simplify: tan27+tan18/1-tan27tan18=
tan(27+18)=tan45
=1
Tori Holiday Blog #2
5 down, 2 to go!
SOHCAHTOA.
sin(theta)=opposite/hypotenuse
cos(theta)=adjacent/hypotenuse
tan(theta)=opposite/adjacent
you can find all the parts of the triangle with just the ones on top, but if they ask for something specific like csc, sec, or cot,you go by this
csc(theta)=hypotenuse/opposite
sec(theta)=hypotenuse/adjacent
cot(theta)=adjacent/opposite
this will only work if you are given a RIGHT triangle, where the hypotenuse will be opposite of the right angle.
When you are given an angle and the right angle, you will go and find the other angle inside the triangle, but you should still always use the one you are given incase you have made a mistake, because if you made a mistake on that, your whole problem will be WRONG!
SOHCAHTOA.
sin(theta)=opposite/hypotenuse
cos(theta)=adjacent/hypotenuse
tan(theta)=opposite/adjacent
you can find all the parts of the triangle with just the ones on top, but if they ask for something specific like csc, sec, or cot,you go by this
csc(theta)=hypotenuse/opposite
sec(theta)=hypotenuse/adjacent
cot(theta)=adjacent/opposite
this will only work if you are given a RIGHT triangle, where the hypotenuse will be opposite of the right angle.
When you are given an angle and the right angle, you will go and find the other angle inside the triangle, but you should still always use the one you are given incase you have made a mistake, because if you made a mistake on that, your whole problem will be WRONG!
Tori's Holiday Blog #1
4 down, 3 to go!
To find a reference angle for sin and cos you must follows these steps:
step 1: figure out the which quadrant the angle is in. (if given a fraction with pi, use pi as 180 and find your angle)
step 2: determine whether the function is positive or negative.
step 3: Subtract 180 from the angle measure until it is between 0 and 90 degrees.
step 4: If the angle comes out to be 0, 30, 45, 60, or 90 degrees, plug it into the trig chart. If it is not on the trig chart then leave it as is or plug it into your calculator.
(*HINT: 0 degrees= 0; 30 degrees- pi/6; 45 degrees- pi/4; 60 degrees- pi/3; 90 degrees- pi/2*
Ex:Find the reference angle for sin 6pi/5
sin (6 x 180)/5 = 1080/5 = 216(treat pi as 180)-sin 216
this angle would be in the III quadrant because it is between 180 and 270
it would also be negative because y is negative in this quadrant.
216-180= 36(subtract from 180)
So, your final answer would be -sin36 degrees.
To find a reference angle for sin and cos you must follows these steps:
step 1: figure out the which quadrant the angle is in. (if given a fraction with pi, use pi as 180 and find your angle)
step 2: determine whether the function is positive or negative.
step 3: Subtract 180 from the angle measure until it is between 0 and 90 degrees.
step 4: If the angle comes out to be 0, 30, 45, 60, or 90 degrees, plug it into the trig chart. If it is not on the trig chart then leave it as is or plug it into your calculator.
(*HINT: 0 degrees= 0; 30 degrees- pi/6; 45 degrees- pi/4; 60 degrees- pi/3; 90 degrees- pi/2*
Ex:Find the reference angle for sin 6pi/5
sin (6 x 180)/5 = 1080/5 = 216(treat pi as 180)-sin 216
this angle would be in the III quadrant because it is between 180 and 270
it would also be negative because y is negative in this quadrant.
216-180= 36(subtract from 180)
So, your final answer would be -sin36 degrees.
kaitlyn's holiday blog 4
chapter 7-4!!!!
this chapter deals with finding reference angles for different functions.
steps: 1)figure out which quadrant the angle is in
2)determine whether or not the function is positive or negative
3)subtract 180 from the angle until it is between 0 and 90
4)once you do that, either use your trig chart or a calculator to find the answer.
Example: find the reference angle for sin420
-this is in quadrant 1
-sin is positive in quadrant 1
-if you keep subtracting 180 from 420, you will get 60
-sin60=(squarerootof3/2) <---answerrrrr
Example: find the reference angle for cos210
-thsi is in quadrant 3
-cos is negative in quadrant 3
-if you keep subtracting 180 from 210, you will get 30
-cos30=(-squarerootof3/2) <---answerrrrr
this chapter deals with finding reference angles for different functions.
steps: 1)figure out which quadrant the angle is in
2)determine whether or not the function is positive or negative
3)subtract 180 from the angle until it is between 0 and 90
4)once you do that, either use your trig chart or a calculator to find the answer.
Example: find the reference angle for sin420
-this is in quadrant 1
-sin is positive in quadrant 1
-if you keep subtracting 180 from 420, you will get 60
-sin60=(squarerootof3/2) <---answerrrrr
Example: find the reference angle for cos210
-thsi is in quadrant 3
-cos is negative in quadrant 3
-if you keep subtracting 180 from 210, you will get 30
-cos30=(-squarerootof3/2) <---answerrrrr
Nathan's last holiday blog
For this blog, I will review on how to get co-terminal angles in degrees and radians. This was another very easy section.
When trying to find a co-terminal angle, you must add or subtract 360, or if you are in radians, add or subtract 2Ï€.
Find a positive co-terminal angle for 27.
All you have to do is add 27+360=387.
Find a negative co-terminal angle for 18.
All you have to do is subtract 18-360=-342
Find a positive co-terminal angle for 2Ï€/3.
All you have to do is add 2Ï€/3+2Ï€=8Ï€/3
Find a negative co-terminal angle for π/2.
All you have to do is subtract π/2-2π=-3π/2
To find a reference angle you must add or subtract 180 until the angle is between the absolute value of 0 and 90.
Find a reference angle for sin 225 degrees.
Just subtract 225-180= sin45=√2/2
Find a reference angle for cos 180 degrees.
Just subtract 180-180= cos0=1
That's all for this blog.
When trying to find a co-terminal angle, you must add or subtract 360, or if you are in radians, add or subtract 2Ï€.
Find a positive co-terminal angle for 27.
All you have to do is add 27+360=387.
Find a negative co-terminal angle for 18.
All you have to do is subtract 18-360=-342
Find a positive co-terminal angle for 2Ï€/3.
All you have to do is add 2Ï€/3+2Ï€=8Ï€/3
Find a negative co-terminal angle for π/2.
All you have to do is subtract π/2-2π=-3π/2
To find a reference angle you must add or subtract 180 until the angle is between the absolute value of 0 and 90.
Find a reference angle for sin 225 degrees.
Just subtract 225-180= sin45=√2/2
Find a reference angle for cos 180 degrees.
Just subtract 180-180= cos0=1
That's all for this blog.
lawrence's holiday blog 4
in this blog i will talk about chapter 13 section 1. in this chapter it deals with arithmatic and geometric. artithmatic is when you add the number to every term. geometric is when you multiply by the same nymber for every term. all of this has to do with sequences which is just a list of numbers,
examples:
1) 3,7,11,15....... find a formula for the Nth term
tn= t1+ (n-1)d
tn= 3+(n-1)4
tn= 3+ 4n-1
tn= 4n-1
2) find the formula for the Nth term of 3
3, 9/2, 27/4
9/2 all over 3/1= 9/6= 3/2
27/4 all over 9/2= 3/2
tn= 3x(3/2) n-1
this is all that i learned in section 1 of chapter 13. it is really easy and if anyone needs help you can just ask me. well now im going do my math homework.
examples:
1) 3,7,11,15....... find a formula for the Nth term
tn= t1+ (n-1)d
tn= 3+(n-1)4
tn= 3+ 4n-1
tn= 4n-1
2) find the formula for the Nth term of 3
3, 9/2, 27/4
9/2 all over 3/1= 9/6= 3/2
27/4 all over 9/2= 3/2
tn= 3x(3/2) n-1
this is all that i learned in section 1 of chapter 13. it is really easy and if anyone needs help you can just ask me. well now im going do my math homework.
lawrence's holiday blog 3
Converting degrees to rads and rads to degrees.
To convert degrees to rads, its degree X pi/180
Ex. 45 degrees
45 X pi/180= 180/6= pi/4.
Ex. for converting rads to degrees.
pi/6
pi/6 X 180/pi= 180/6 = 30 degrees
remember that the pi's cancel out!!!
I also understood how to get positive and negative coterminal angles.
if its in degrees you substract or add from 360. if its a large negative number and you need to get a positive coterminal angle keep adding it from 360 till you get a positive number.
ex. -1030 degrees
-1030+360+360+360= 50 degrees. (positive)
-1030-360= -1390 degrees. (negative)
if its in radians you just add 2pi or subtract 2pi
Ex. pi/6
pi/6+ 2pi= 13pi/6 (positive)
pi/6- 2pi= -11pi/6 (negative)
The other concept i understood was from 7-4
How to find a reference angle.
1. find the quadrant the angle is in
2. determine if the trig function is positive or negative
3. subtract 180 degrees from the angle until theta is b/w 0 and 90 degrees
4. if it is a trig chart angle plug in. if not leave it or plug in calculator.
ex. cos 225 degrees
its in quad 3
its negative (-cos)
225-180= 45 -cos 45 degrees. (trig chart function)
= -square root 2/2
that is all the concepts that i know the best the others i am decent with so if you need help ask me in class and i will be glad to help.
To convert degrees to rads, its degree X pi/180
Ex. 45 degrees
45 X pi/180= 180/6= pi/4.
Ex. for converting rads to degrees.
pi/6
pi/6 X 180/pi= 180/6 = 30 degrees
remember that the pi's cancel out!!!
I also understood how to get positive and negative coterminal angles.
if its in degrees you substract or add from 360. if its a large negative number and you need to get a positive coterminal angle keep adding it from 360 till you get a positive number.
ex. -1030 degrees
-1030+360+360+360= 50 degrees. (positive)
-1030-360= -1390 degrees. (negative)
if its in radians you just add 2pi or subtract 2pi
Ex. pi/6
pi/6+ 2pi= 13pi/6 (positive)
pi/6- 2pi= -11pi/6 (negative)
The other concept i understood was from 7-4
How to find a reference angle.
1. find the quadrant the angle is in
2. determine if the trig function is positive or negative
3. subtract 180 degrees from the angle until theta is b/w 0 and 90 degrees
4. if it is a trig chart angle plug in. if not leave it or plug in calculator.
ex. cos 225 degrees
its in quad 3
its negative (-cos)
225-180= 45 -cos 45 degrees. (trig chart function)
= -square root 2/2
that is all the concepts that i know the best the others i am decent with so if you need help ask me in class and i will be glad to help.
Saturday, January 1, 2011
kaitlyn's holiday blog 3
chapter 13-1!!!!
this section was superr easy because it deals with patterns of different sets of numbers.
Arithmatic-you add to get the next term in an arithmatic sequence
Geometric-you multiply to get the next term in a geometic sequence
Examples: is this sequence arithmatic or geometric? 2,5,8,11
arithmatic, because you add 3 to each numberrr.
Examples: is this sequence arithmatic or geometric? 4,8,16,32
geometric, because you mulitply by 2 to each numberrr.
Examples: find the next to numbers in this arithmatic sequence; 4,7,10,13
16,19 <---answerrrrrr
Examples: find the next two numbers in this geometric sequence; 3,9,27
81,243 <---answerrrrr
this section was superr easy because it deals with patterns of different sets of numbers.
Arithmatic-you add to get the next term in an arithmatic sequence
Geometric-you multiply to get the next term in a geometic sequence
Examples: is this sequence arithmatic or geometric? 2,5,8,11
arithmatic, because you add 3 to each numberrr.
Examples: is this sequence arithmatic or geometric? 4,8,16,32
geometric, because you mulitply by 2 to each numberrr.
Examples: find the next to numbers in this arithmatic sequence; 4,7,10,13
16,19 <---answerrrrrr
Examples: find the next two numbers in this geometric sequence; 3,9,27
81,243 <---answerrrrr
kaitlyn's holiday blog 2
chapter 9-1!!!!
in this chapter, we learned how to solvee right angles using SOHCAHTOA. this is a shortcut you use to find the sides of the triangles and even the angles.
S-sine
O-opposite
H-hypoteneuse
C-cosine
A-adjacent
H-hypoteneuse
T-tangent
O-opposite
A-adjacent
Examples: A=90; B=30; c=5; a=13
-first you want to find C. in order to do this you must use SOHCAHTOA.
use sinx=5/13
C=60
-then you want to find b. use sin30=x/13
b=12
in this chapter, we learned how to solvee right angles using SOHCAHTOA. this is a shortcut you use to find the sides of the triangles and even the angles.
S-sine
O-opposite
H-hypoteneuse
C-cosine
A-adjacent
H-hypoteneuse
T-tangent
O-opposite
A-adjacent
Examples: A=90; B=30; c=5; a=13
-first you want to find C. in order to do this you must use SOHCAHTOA.
use sinx=5/13
C=60
-then you want to find b. use sin30=x/13
b=12
Kaitlyn's holiday blog 1
chapter 10-1!!!!
this was probably one of the easiest chapters for me, all you need to know is the two formulas and your trig chart and you will be fine.
formulas: sin(A+-B)=sinAcosB+-cosAsinB
cos(A+-B)=cosAcosB-+sinAsinB
*A and B represent different angle numbers.
Examples: find sin75
-sin(45+30)=sin45cos30+cos45sin30
-sin(75)=(squarerootof 2/2)(squarerootof 3/2)+(squarerootof 2/2)(1/2)
-sin(75)=(squarerootof 6/4)+(squarerootof 2/4)
-(squarerootof 6)+(squarerootof 2)/4 <---answerrrrrr
Examples: find cos30
-cos(60-30)=cos60cos30+sin60sin30
-cos(30)=(1/2)(squarerootof 3/2)+(squarerootof 3/2)(1/2)
-cos(30)=(squarerootof 3/4)+(squarerootof 3/4)
-(squarerootof 6/4) <----answerrrrrrr
Examples: simplifyyy sin60cos30+cos60sin30
-sin(60+30)
-sin(90)
- 1 <----answerrrr
this was probably one of the easiest chapters for me, all you need to know is the two formulas and your trig chart and you will be fine.
formulas: sin(A+-B)=sinAcosB+-cosAsinB
cos(A+-B)=cosAcosB-+sinAsinB
*A and B represent different angle numbers.
Examples: find sin75
-sin(45+30)=sin45cos30+cos45sin30
-sin(75)=(squarerootof 2/2)(squarerootof 3/2)+(squarerootof 2/2)(1/2)
-sin(75)=(squarerootof 6/4)+(squarerootof 2/4)
-(squarerootof 6)+(squarerootof 2)/4 <---answerrrrrr
Examples: find cos30
-cos(60-30)=cos60cos30+sin60sin30
-cos(30)=(1/2)(squarerootof 3/2)+(squarerootof 3/2)(1/2)
-cos(30)=(squarerootof 3/4)+(squarerootof 3/4)
-(squarerootof 6/4) <----answerrrrrrr
Examples: simplifyyy sin60cos30+cos60sin30
-sin(60+30)
-sin(90)
- 1 <----answerrrr
Mary's Last holiday blog! woo!
7-4
The steps:
1. Find the Quadrant
2. Determine + or
3. Subtract 180 until b/w 0 & 90
4. Use trig chart or leave it.
Detailed Example: Find a reference anlge of cos225 degreesFirst, find the quadrant, 225 is between 180 and 270 so its in the third quadrant.Cos is related to the x axis, x is negative in the third quadrant, which makes it negativeNext, start subtracting 180 until you reach an angle between 0 and 90225-180=45 degrees, which is on the trig chart! joy!45 degrees is pi/4, take out your trig chart and look at the cos part, look for pi/4there's your answer! -squareroot of 2/2 (the negative is from the second step,it is carried)
Short Example: Find a reference angle of sin 600 degrees
1. Quadrant III
2. Negative
3. 60 degrees
4. 60 = pi/3 on trig chart
Final answer: Sin600=-sin60=square root of 3/2
Mary's holiday blog 3
section 13-2
this section is just about recorsive numbers in a sequence
tn-1 means previous term
tn-2 means two terms back
tn-3 means three terms back
recorsive means define in terms of what came before
find the 2nd and 3rd terms
they give you a starting number and a sequence
t1=7 tn=tn-1+1
brob said you look at tn-1 as the term before and not as anything minus 1
so you take your first number and you plug it in
you will get tn=4(7)+1
then you get 29
so then you plug that number in to get the next one
so you get tn=4(29)+1 which equals 117
you do that to find infinite numbers in a recorsive sequence
example number 2
they could ask you what was asked above or they could give you the numbers and ask you to write a recorsive definition. which means a formula!
so if they gave you 4,7,10,13
you can see that they are adding three
soo tn=tn-1=3
and you use that formula to find other numbers
Mary's holiday blog 2
Review of 10-2
You can't do anything unless you know these formulas:
tan(a+b)=tan(a)+tan(b)/1-tan(a)tan(b)
tan(a-b)=tan(a)-tan(b)/1+tan(a)tan(b)
they could tell you something like "suppose tan(.." and you will find an answer by using the full formula, if they tell you to simplify, you will use the opposite side of the equation.
example:
simplify
tan27+tan18/1-tan27tan18
so you know you need to use the other side of the formula
which is tan(a +b)
which is tan (45) which = 1
yay
math is fun...
example 2: tan( x + pie/4)
so you find the formula that matches
so you plug your numbers in.
so you get
tanx+tanpie/4/1-tanxtanpie/4
which simplifies to,
tanx+1/1-tanx
or they could tell you this
"suppose tan alpha=1/3 tan alpha beta=1/2
and then you'll put that straight into the formula and you get:
1/3 + 3/6/1-1/6=1
but if it had said cos alpha = 1/3, you would have to draw out your triangle and take the tan of it.
yeaa!
Mary's holiday blog 1
Review of 9-1
An easy way to remember these is SOHCAHTOA.
sin(theta)=opposite/hypotenuse
cos(theta)=adjacent/hypotenuse
tan(theta)=opposite/adjacent
you can find all the parts of the triangle with just the ones on top, but if they ask for something specific like csc, sec, or cot,you go by this
csc(theta)=hypotenuse/opposite
sec(theta)=hypotenuse/adjacent
cot(theta)=adjacent/opposite
this will only work if you are given a RIGHT triangle, where the hypotenuse will be opposite of the right angle.
When you are given an angle and the right angle, you will go and find the other angle inside the triangle, but you should still always use the one you are given incase you have made a mistake, because if you made a mistake on that, your whole problem will be WRONG!
Example.
A=90 C=65 and a=18
you draw your triangle and use 180-90-65, to get your B angle.
then you use cos to find c
and you use sin, to get b
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