Friday, August 12, 2011
Sunday, April 3, 2011
Dom's blog
I don't know about those guys but i thought this week was pretty awesome, we got to get credit for easy tests every day, catch up on our aleks, and have some fun. My favorite was probably the chapter seven test because it was the easiest one to do. I'm surprised I still remember that trig chart. Well thats it...peace.
Monday, March 28, 2011
Nathan's blog
This week in advanced math, we basically did a bunch of aleks. Monday we had SAT's and everyone else went on a field trip. Then the rest of the week we started on chapter 6, and on thursday and friday, we did aleks with the sub. Chapter 6 had to deal with equations of circles and ellipses, and stuff like that. I checked out on wednesday and missed a little bit of the lesson but i should be able to pick up on it pretty fast. That is all for this blog.
Sunday, March 27, 2011
mary'ss blog
this. week. was. crazy. We had to sell candy and get ready for state, and we did aleks, alot of them, and then we couldn't do work because brob left us. haha. But dang, all them aleks really helped during everything i did at state, especially the game descartes. i swear i wouldn't have known half the stuff i knew without it. i really like math club, it's really fun and worth it, i can't even imagine not being in it at this point:)) this question came up on a 10 point card in how many zeros does 127! end! ahahahah!!!!!!!!!!1
Nicala's Post
this week we did not really get to do anything. because of the sat and the field trip and then thing with mu alpha theta. so we mainly did aleks the whole week. Which most of them i did not know what i am doing so i get put out or i fail my assestment and have to do the same things over and over. okay so i am going to back to chapter 13 section 1. In this chapter they are talking about sequences. A sequences is a set of number. There are two different type of sequences we learn about in this section, arithmetic and geometric. I am only talking about arithmetic because i am sleeping and ready to go to bed lol :) Arithmetic squence is when you add the same number to every term. The only formula you need to know for this problem is tn=t1+(n-1)d. d is the number that you are adding to each term. Example Problem Find the first four terms of this equation 7n+2. in the problem you always replace n with number term you are on. t1=7(1)+2=9 t2=7(2)+2=16 t3=7(3)+2=23 t4=7(4)+2=30 So the first four terms in this equation is 9,16,23,30
Lawrence's blog
this week was really crazy. we couldnt really do much in the class because of the SAT and people that were in mu alpha theta. so really all we did was Aleks. i didnt know how to do any of the ones that we had to do. i attempted even copied the problems and tried workign them step for step and still couldnt get them. so i guess im screwed till it makes me take a dumb 30 question test. if there is anyone who is below me and may need my help i can help you with it and hopefully for some odd reason someone will know how to do the stuff im on
Friday, March 25, 2011
Another One Of Taylor's Blogs
The following is part one of my notes on lesson 6.3.
In this lesson we will be doing the easier parts of finding the parts of the equation.
The one we will be doing are in this form
x^2/a+y^2/b=1
We will by finding the minor and major axis and the length of the major and minor axis.
Equations
Length of major 2 square root of lager denom.
Length of minor 2 square root of smaller denom.
Lets do some example problems
Find the minor and major axis and the length of the major and minor axis of x^2/9+y^2/25=1
First we will identify the major and minor axis. In this case the major axis is y and the minor axis is x.
2 square root of 25= 2 x 5= 10 Then we simply plug the major axis’s denominator into this equation and solve it to find the length of the major axis.
2 square root of 9= 2 x 3= 6 Then we simply plug the minor axis’s denominator into this equation and solve it to find the length of the minor axis.
And your answers will be major axis= y, minor axis= x, length of the major axis= 10, length of the minor axis= 6. Lets do one more!
Find the minor and major axis and the length of the major and minor axis of x^2/81+y^2/36=1
First we will identify the major and minor axis. In this case the major axis is x and the minor axis is y.
2 square root of 81= 2 x 9= 18 Then we simply plug the major axis’s denominator into this equation and solve it to find the length of the major axis.
2 square root of 36= 2 x 6= 12 Then we simply plug the minor axis’s denominator into this equation and solve it to find the length of the minor axis.
And your answers will be major axis= x, minor axis= y, length of the major axis= 18, length of the minor axis= 12.
THAT IS ALL FOR THIS PART OF MY LESSON 6.3 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
In this lesson we will be doing the easier parts of finding the parts of the equation.
The one we will be doing are in this form
x^2/a+y^2/b=1
We will by finding the minor and major axis and the length of the major and minor axis.
Equations
Length of major 2 square root of lager denom.
Length of minor 2 square root of smaller denom.
Lets do some example problems
Find the minor and major axis and the length of the major and minor axis of x^2/9+y^2/25=1
First we will identify the major and minor axis. In this case the major axis is y and the minor axis is x.
2 square root of 25= 2 x 5= 10 Then we simply plug the major axis’s denominator into this equation and solve it to find the length of the major axis.
2 square root of 9= 2 x 3= 6 Then we simply plug the minor axis’s denominator into this equation and solve it to find the length of the minor axis.
And your answers will be major axis= y, minor axis= x, length of the major axis= 10, length of the minor axis= 6. Lets do one more!
Find the minor and major axis and the length of the major and minor axis of x^2/81+y^2/36=1
First we will identify the major and minor axis. In this case the major axis is x and the minor axis is y.
2 square root of 81= 2 x 9= 18 Then we simply plug the major axis’s denominator into this equation and solve it to find the length of the major axis.
2 square root of 36= 2 x 6= 12 Then we simply plug the minor axis’s denominator into this equation and solve it to find the length of the minor axis.
And your answers will be major axis= x, minor axis= y, length of the major axis= 18, length of the minor axis= 12.
THAT IS ALL FOR THIS PART OF MY LESSON 6.3 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
Monday, March 21, 2011
Charlie.
Ok, so we didn't have any aleks due this week.
BUT, we did do stuff for math this week.
we learned chapter 6.
one thing we did was the eclipses & stuff like that.
the formula is: x^2/a + y^2/b = 1
to do this we have to find the major axis, the minor axis, the major length, the minor length, the vertex, other, and the focus.
EXAMPLE:
x^2/4 + y^2/16 = 1
Major --> y
Minor --> x
Major Length --> 8
Minor Length --> 4
Vertex --> (0, 4) (0, -4)
Other --> (2, 0) (-2, 0)
Focus --> (0, 2squarerootof3) (0, -2squarerootof3)
BUT, we did do stuff for math this week.
we learned chapter 6.
one thing we did was the eclipses & stuff like that.
the formula is: x^2/a + y^2/b = 1
to do this we have to find the major axis, the minor axis, the major length, the minor length, the vertex, other, and the focus.
EXAMPLE:
x^2/4 + y^2/16 = 1
Major --> y
Minor --> x
Major Length --> 8
Minor Length --> 4
Vertex --> (0, 4) (0, -4)
Other --> (2, 0) (-2, 0)
Focus --> (0, 2squarerootof3) (0, -2squarerootof3)
Sunday, March 20, 2011
Feroz's Blog
Haven't done one of these in awhile. Anyway, I'm gonna cover Chapter 6.
Chapter 6 deals with circles, parabolas, hyperboles and ellipses and stuff.
ex. x^2/16 + y^2/25 = 1
1. Major axis = variable with the greater denom = y
2. Minor axis = variable with the smaller denom = x
3. Length of Major = 2 x sqrt(major) = 10
4. Length of Minor = 2 x sqrt(minor) = 8
5. Vertex = since y is the major = (0,5), (0,-5)
6. Other int. = same thing with the minor = (4,0), (-4,0)
7. Focus = sqrt(major - minor) = sqrt(2) = (0, sqrt(2)), (0, -sqrt(2))
8. Sketch the graph
There you go.
Nathan's blog
This week, we started on chapter 6. It has to deal with circles, ellipses, and hyperbolas. So far it is pretty easy, cause the formulas and equations don't differ that much. Each one has a seven step process, and this basically helps you find all of the parts of the shape, and helps with plugging into the calculator, so that you can sketch the shape. Here's an example of finding the center and radius of a circle.
Ex. (x-3)^2 + (y+7)^2 = 19
C: (3,-7)
r= √19
Find all the parts of: x^2/4 + y^2/25 = 1
1. Major axis: y
2. Minor axis: x
3. Length of major: 2√25=10
4. Length of minor: 2√4=4
5. Vertex: (0,5) (0,-5)
6. Other intercept: (2,0) (-2,0)
7. Focus: 4=25-f^2 -21=-f^2 f=√21 (0,√21) (0,-√21)
That's about it for this review. The only difference in the other formulas, is that you have to find aymptotes. Otherwise, this chapter is very easy so far.
Ex. (x-3)^2 + (y+7)^2 = 19
C: (3,-7)
r= √19
Find all the parts of: x^2/4 + y^2/25 = 1
1. Major axis: y
2. Minor axis: x
3. Length of major: 2√25=10
4. Length of minor: 2√4=4
5. Vertex: (0,5) (0,-5)
6. Other intercept: (2,0) (-2,0)
7. Focus: 4=25-f^2 -21=-f^2 f=√21 (0,√21) (0,-√21)
That's about it for this review. The only difference in the other formulas, is that you have to find aymptotes. Otherwise, this chapter is very easy so far.
BLLOOGGG!
i love getting grades for blogs:)
thiiisss weeeeeeeeeek.
We are learning about chapter 6 about finding all the parts of a hyperbola and an ellipse.
there isn't really much to explain about it so i'll just walk you through the example.
Ex. 1 which is an ellipse, and you can tell the difference between and ellipse and a hyperbola by looking to see if there is a negative.
x^2/36 + y^2/49 =1 and to be in standard form, it will always have to equal one.
so to start it off
Major axis-y-because the bigger number is under the 7
minor axis-x- by default
vertex-(0,7),(0,-7) because of the major axis which they follow
focus- focus^2=bigger plus smaller=(0,plus or minus-square root of 85)-still using the place of the major axis
other intercept( which you use to sketch a graph)-(6,0)(-6,0) you use the square root of the minor axis in that place.
length of major axis-14
length of minor-12, you can get this by doubling the square root of the number under the x or y.
and the process is really close for a hyperbola, but there is the negative to deal with and you have to find asyptotes. but its really easy, everybody should really try really hard in this last semester and finish out strong and not have to worry about next year.::))
Nicala's Post
In class we are working on chapter six we did three sections already. the sections we worked on are circles, ellipses, and hyperbolas. Ellipses and hyperbolas are basically the same expect the change in their formulas and hyperbolas have asymptotes and ellipses do not asymptotes. Ellipses have a oval like shape. The formula for its
x^2/a^2+y^2/b^2=1
The formula for an hyperbola is the same except the y or x is negative and that changes the formula. When your working with ellipses you are looking for the major, minor, length of major, length of minor, vertex, other intercepts, and the focus and depending on the instructions you graph.
example problem
x^2/4+y^2/16=1
1.major-y because 16 is bigger than 4
2.minor is x because 4 is smaller than 16.
3. length of major: 2 the square root of 16= 8
4. length of minor: 2 the square root of 4=4
5.vertex is the square to the major and put the answer in points= (0,4) &(0,-4)
6. the other intercept is the square root of the minor and put the answer in points= (2,0) &(-2,0)
7.focus is the both denominators equal to f squared. 4-16=f^2 and now solve like normal and it gives u an answer of f=2 the square root of 3 and you put your final in points. (0, 2 the square root of 3) & (0,-2 the square root of 3)
8. sketching the graph is option
bloggggg
This weekk, we started learning chapter 6. this chapter is pretty much about graphs like hyperboles, ellipes, and circles.
Circle
equation: (x-h)^2+(y-k)^2=r^2
where (h,k) is center
To put an equation into the standard form of a circle, you have to complete the square.
Ellipse
equation: x^2/a + y^2/b=1
major axis: variable with larger denomenator
minor axis: varibale with smaller denomenator
length of major: 2(squarerootof larger)
length of minor: 2(squarerootof smaller)
vertex: (squarerootof larger) if x is major: ( , 0) (- , 0)
if y is major: (0, ) (0,- )
other int. : (squarerootof smaller) ( , )
focus: smaller=larger-focus^2
Hyperbole
This is pretty much the same as the ellipse, except you also have to deal with negative numbers and asymptotes. also the focus is different.
focus: focus^2=larger+smaller
asymptote: y=+-b/a x if x is major
y=+-a/b x if y is major
Circle
equation: (x-h)^2+(y-k)^2=r^2
where (h,k) is center
To put an equation into the standard form of a circle, you have to complete the square.
Ellipse
equation: x^2/a + y^2/b=1
major axis: variable with larger denomenator
minor axis: varibale with smaller denomenator
length of major: 2(squarerootof larger)
length of minor: 2(squarerootof smaller)
vertex: (squarerootof larger) if x is major: ( , 0) (- , 0)
if y is major: (0, ) (0,- )
other int. : (squarerootof smaller) ( , )
focus: smaller=larger-focus^2
Hyperbole
This is pretty much the same as the ellipse, except you also have to deal with negative numbers and asymptotes. also the focus is different.
focus: focus^2=larger+smaller
asymptote: y=+-b/a x if x is major
y=+-a/b x if y is major
Friday, March 18, 2011
Taylor's Mystery Number Blog (I GIVE UP ON NUMBERING IT T-T!!)
The following is part one of my notes on lesson 6.2.
In this lesson we will be learning a few new things about circles.
You will need these equations in order to do this properly:
Distance formula= square root (y2-y1)^2+(x2+x1)^2= radius
The 2 and ones are small and are at the bottom to symbolize that they are from either the first and second point . One of these points must be the center while the other must be on the circle.
Equation of a circle (x-h)^2+(y-k)^2=r^2 the (h,k) is the center of the circle.
Now lets practice how to find the radius and the center of the circle from the equation of the circle. Which is also the easiest and simplest part to do for this chapter.
Find the center and the radius of (x-9)^2+(y-8)^2=16.
From the equation of the circle we can see that the place where 9 and 8 are is replacing the h and the k so that means that the center of the circle is (9,8).
We also know that the radius is raised to the second before it can be considered part of the equation so we just need to find the square root of 16 in order to find the radius. So that makes the radius 4.
Find the center and the radius of (x+5)^2+(y+11)^2=15.
From the equation of the circle we can see that the place where 5 and 11 are is replacing the h and the k but since they are not negative we have to make them negative when we write it in coordinate form. So the center of the circle is (-5,-11)
We also know that the radius is raised to the second before it can be considered part of the equation so we just need to find the square root of 15 in order to find the radius but since we cannot we will simply leave it with the square root as the answer. So that makes the radius square root of 15.
THAT IS ALL FOR THIS PART OF MY LESSON 6.2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
In this lesson we will be learning a few new things about circles.
You will need these equations in order to do this properly:
Distance formula= square root (y2-y1)^2+(x2+x1)^2= radius
The 2 and ones are small and are at the bottom to symbolize that they are from either the first and second point . One of these points must be the center while the other must be on the circle.
Equation of a circle (x-h)^2+(y-k)^2=r^2 the (h,k) is the center of the circle.
Now lets practice how to find the radius and the center of the circle from the equation of the circle. Which is also the easiest and simplest part to do for this chapter.
Find the center and the radius of (x-9)^2+(y-8)^2=16.
From the equation of the circle we can see that the place where 9 and 8 are is replacing the h and the k so that means that the center of the circle is (9,8).
We also know that the radius is raised to the second before it can be considered part of the equation so we just need to find the square root of 16 in order to find the radius. So that makes the radius 4.
Find the center and the radius of (x+5)^2+(y+11)^2=15.
From the equation of the circle we can see that the place where 5 and 11 are is replacing the h and the k but since they are not negative we have to make them negative when we write it in coordinate form. So the center of the circle is (-5,-11)
We also know that the radius is raised to the second before it can be considered part of the equation so we just need to find the square root of 15 in order to find the radius but since we cannot we will simply leave it with the square root as the answer. So that makes the radius square root of 15.
THAT IS ALL FOR THIS PART OF MY LESSON 6.2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
Tuesday, March 15, 2011
Charlie
this week we had the holidays..
umm, i did 39 of my 50 bonus point modules.
my computer was broken until wednesday so i did it thursday and friday.
& then a couple more on sunday. i hope not dong this doesn't make me fail. but whateva.
i don't remember what we were learning before we went into the holidays.
but on aleks i didn't know how to do the blue, purple, or yellow stuff.
so i went do some of the trig. stuff & what i had unlocked for the matrices thing.
but i still didn't know how to do anything else to get to my 50 modules.
so that isn't fair, JUS SAYin.
it's hot in this class.
umm, i did 39 of my 50 bonus point modules.
my computer was broken until wednesday so i did it thursday and friday.
& then a couple more on sunday. i hope not dong this doesn't make me fail. but whateva.
i don't remember what we were learning before we went into the holidays.
but on aleks i didn't know how to do the blue, purple, or yellow stuff.
so i went do some of the trig. stuff & what i had unlocked for the matrices thing.
but i still didn't know how to do anything else to get to my 50 modules.
so that isn't fair, JUS SAYin.
it's hot in this class.
Sunday, March 13, 2011
Nathan's blog
Yeah, so i didnt do my aleks and now I'm mad, so i might as well get some points for this. I'm probably just going to do a basic review of stuff we learned in class, and stuff in aleks.
So, I'm going to start off by reviewing matrices, which was very easy.
[2 4] [8 3] [10 7]
[7 5] + [9 12] = [16 17]
Now on to subtracting matrices.
[18 6] [5 7] [13 -1]
[15 3] - [14 9] = [ 1 -6]
Now, i will review how to find the slope of a line.
(4,3) (7,2)
2-4/7-3
The slope of the line is: -1/2
The slope of a line is 4/3. Find the slope of a parallel and perpendicular line, using this slope.
For the slope of a parallel line, it is the same as the slope of the original line. So the slope of this parallel line is: 4/3
When finding the slope of a perpendicular line, you take the negative reciporcal of the original slope. So the slope of this perpendicular line is: -3/4
That's it for this review.
So, I'm going to start off by reviewing matrices, which was very easy.
[2 4] [8 3] [10 7]
[7 5] + [9 12] = [16 17]
Now on to subtracting matrices.
[18 6] [5 7] [13 -1]
[15 3] - [14 9] = [ 1 -6]
Now, i will review how to find the slope of a line.
(4,3) (7,2)
2-4/7-3
The slope of the line is: -1/2
The slope of a line is 4/3. Find the slope of a parallel and perpendicular line, using this slope.
For the slope of a parallel line, it is the same as the slope of the original line. So the slope of this parallel line is: 4/3
When finding the slope of a perpendicular line, you take the negative reciporcal of the original slope. So the slope of this perpendicular line is: -3/4
That's it for this review.
Friday, March 11, 2011
Taylor's 14th(I think XD) Blog Review
REVIEW TIME!!!!
This is the review on my notes from Ch. 13.2.
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=4 tn=8tn-1+7
t1=8 (4)+7= 39 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 8 (39) + 7 = 319
t3= 8 (319) + 7= 2559
t4= 8 (2559) + 7 = 20479
t5= 8 (20479) + 7 = 163839
So your answers will be 2559, 20479, and 163839.
Now let's try finding the recursive definition for 8, 12, 16, 20...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 4 in between each number your answer or equation will be tn-1+4.
This is the review on the rest of my notes on Ch. 13.2. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
This is the review on my notes from Ch. 13.2.
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=4 tn=8tn-1+7
t1=8 (4)+7= 39 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 8 (39) + 7 = 319
t3= 8 (319) + 7= 2559
t4= 8 (2559) + 7 = 20479
t5= 8 (20479) + 7 = 163839
So your answers will be 2559, 20479, and 163839.
Now let's try finding the recursive definition for 8, 12, 16, 20...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 4 in between each number your answer or equation will be tn-1+4.
This is the review on the rest of my notes on Ch. 13.2. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
Sunday, March 6, 2011
Nathan's blog
So, we just took our exams on chapter 5 which dealt with logs and interest, and a few more formulas. I thought I understood the information, but apparently not. I just finished the 14 aleks modules that we needed done by today, and then tommorrow i will start on the 50 modules for extra credit.
So, now i will just review different stuff throughout chapter 5.
Rule of 72
72/r%=how long it takes to double
How long will it take to double your money at a 12% interest?
72/12=6
We learned a formula for interest:
A(t)=Ao b^t/k
Ex. A bank advertises that if you open a savings account, you can double your money in twelve years. Find out how much money you will have after 7 years, if you invest $1,000 into your account.
A(t)=1000(2)^7/12=$1498.31
I'm very happy to get out of chapter 5, hopefully i will understood the information better in the 4th nine weeks, because that will probably be the toughest part of the year in math.
So, now i will just review different stuff throughout chapter 5.
Rule of 72
72/r%=how long it takes to double
How long will it take to double your money at a 12% interest?
72/12=6
We learned a formula for interest:
A(t)=Ao b^t/k
Ex. A bank advertises that if you open a savings account, you can double your money in twelve years. Find out how much money you will have after 7 years, if you invest $1,000 into your account.
A(t)=1000(2)^7/12=$1498.31
I'm very happy to get out of chapter 5, hopefully i will understood the information better in the 4th nine weeks, because that will probably be the toughest part of the year in math.
Nicala's Post
okay we had our exams and we learned the rest of chapter five which was about exponents which we learned in aleks and logs which we should have learned in algebra 2 but i did not understand it then and do not understand now. okay we took our exam thursday which i failed horrible and we started doing fourteen aleks friday and will finish them this weekend and tomorrow we start the fifty bonus aleks for the extra fifty points in class which i need. one of things i learned on aleks is how to do translation of a vector.
Example Problem
A vector with an initial point of (-6,-2) and terminal point of (-5,3) is translated so that its initial point is at the origin. Find its new terminal point.
To find the terminal point you have subtract them from each other.
you subtract -5 minus -6. since you are subtracting and the six is negative the six becomes positive which gives you 1 as your first new terminal point.
next you subtract 3 minus -2 since you are subtracting and the two is the two becomes positive
which gives you 5 as your second new terminal point.
the answer is (1,5)
Example Problem
A vector with an initial point of (-6,-2) and terminal point of (-5,3) is translated so that its initial point is at the origin. Find its new terminal point.
To find the terminal point you have subtract them from each other.
you subtract -5 minus -6. since you are subtracting and the six is negative the six becomes positive which gives you 1 as your first new terminal point.
next you subtract 3 minus -2 since you are subtracting and the two is the two becomes positive
which gives you 5 as your second new terminal point.
the answer is (1,5)
Friday, March 4, 2011
Taylor's 14th(I think XD) Blog Review
REVIEW TIME!!!!
This is the review on the rest of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 3 x 2^ n-1
t1=3 x 2^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=3 x 2^ 2-1
t3=3 x 2^ 3-1
t4=3 x 2^ 4-1
t1= 3 Then you have your answers.
t2= 6
t3= 12
t4= 24
9,36,144,576.... find the formula for the nth term.
9,36,144,576.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 4.
tn= t1 x r ^n-1Then write out your equation.
tn= 9 x4 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
This is the review on the rest of my notes on Ch. 13.1. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
This is the review on the rest of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 3 x 2^ n-1
t1=3 x 2^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=3 x 2^ 2-1
t3=3 x 2^ 3-1
t4=3 x 2^ 4-1
t1= 3 Then you have your answers.
t2= 6
t3= 12
t4= 24
9,36,144,576.... find the formula for the nth term.
9,36,144,576.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 4.
tn= t1 x r ^n-1Then write out your equation.
tn= 9 x4 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
This is the review on the rest of my notes on Ch. 13.1. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
Charlie.
This week we had exams.
But anyways, in aleks i ended up having to take the stupid assessment thingy & got knocked down to like 70 in my blue thingy..
It's friday, in first hour.
& i think today in math class we'll be coming to the lab or A1 to do 7 alek modules.
but not the library because the book fair is this week, so yea.
then, on monday we could start on the extra credit thingy.
extra credit this nine weeks is 50 aleks modules.
which, idk how i'm goin to do that because MY COMPUTER'S BROkEN!
but, i guess i'll figure something out because i would like a really good grade this time.
... This week we learned 2 (or 3?) new sections in chapter (i think) 5.
then we pretty much reviewed for the rest of the week and took our exam yesterday (thursday).
when i was doing the practice chapter test in the text book i knew what i was doing,
but when i actually took the exam; i was like "WhAt?!?"
so, (basically) i do believe that i FAILED(:
example of something:
log2~x = 4
2^4 = x
x = 16
But anyways, in aleks i ended up having to take the stupid assessment thingy & got knocked down to like 70 in my blue thingy..
It's friday, in first hour.
& i think today in math class we'll be coming to the lab or A1 to do 7 alek modules.
but not the library because the book fair is this week, so yea.
then, on monday we could start on the extra credit thingy.
extra credit this nine weeks is 50 aleks modules.
which, idk how i'm goin to do that because MY COMPUTER'S BROkEN!
but, i guess i'll figure something out because i would like a really good grade this time.
... This week we learned 2 (or 3?) new sections in chapter (i think) 5.
then we pretty much reviewed for the rest of the week and took our exam yesterday (thursday).
when i was doing the practice chapter test in the text book i knew what i was doing,
but when i actually took the exam; i was like "WhAt?!?"
so, (basically) i do believe that i FAILED(:
example of something:
log2~x = 4
2^4 = x
x = 16
Sunday, February 27, 2011
Feroz's Blog
Blah blah blah blue pie blah blah Aleks blah Chapter 5.
- Rule 72
ex. How long would it take to double your money at 36% interest?
72/36 = 2
- Rational Exponents
ex. x^1/2
= sqrt(x)
Covered two sections without complaining about exams or a bee in my kitchen. I think this went pretty well.
- Rule 72
ex. How long would it take to double your money at 36% interest?
72/36 = 2
- Rational Exponents
ex. x^1/2
= sqrt(x)
Covered two sections without complaining about exams or a bee in my kitchen. I think this went pretty well.
Mary
|
A street that is 222 m long is covered in snow. City workers are using a snowplos to clear the street. A tire on the snowplow has to turn 37 times n traveling the length of the street. What is the diameter of the tire? crazy right?
| Here is the solution, it took me forever to get this module correct! it took me all class! First, we'll find the circumference C of the tire. Then, we'll use C to calculate the diameter d .
We are told that the tire turns
| ||||||||||||||||||||||||||||||||
Lawrence"s blog
this week we pretty much did chapter 12. we did all of the stuff she taught us on aleks so none of it is really new. it was more of a review than anything. All of chapter 12 seems like a review because of aleks wich is a good thing because we get to learn it and understand it and it doesnt seem like greek anymore. chapter 12 is also our exam i think too if i remember correctly. i wont put an examples cuz you can go on aleks and find them or you can look at the one's brob did but if you need help i will be glad to do so
Nathan's blog
So, this week all we did was review the stuff from chapter 5, which was basically an overview of our blue pie in aleks. This stuff was really easy, and this is the only thing the exam is on.
Section 5-2 was on rational exponents.
-The denominator tells you the root. The numerator is the exponent.
Ex. x^3/2= √x^3
In section 5-3, we learned a new formula.
-A(t)=Ao b^t/x
-Ao=starting amount
-b=Doubling, halfing, triplin, etc.
-t=time
-k=how long it takes to get b.
Ex. A bank advertises that if you open a savings account, you can double your money in 12 years. Find out how much money you will have after 7 years, if you invest $1,000.
Ao=1,000 b=2
t=7 k=12
A(t)=1,000(2)^7/12= $1498.31
Rule of 72
72/r%=how long it takes to double.
Ex. How long will it take to double your money at 12% interest?
72/12=6
Chapter 5 is pretty easy, and I should do good on the exam, because I understand most of it due to the fact that it is basically a review of what we've been doing.
Section 5-2 was on rational exponents.
-The denominator tells you the root. The numerator is the exponent.
Ex. x^3/2= √x^3
In section 5-3, we learned a new formula.
-A(t)=Ao b^t/x
-Ao=starting amount
-b=Doubling, halfing, triplin, etc.
-t=time
-k=how long it takes to get b.
Ex. A bank advertises that if you open a savings account, you can double your money in 12 years. Find out how much money you will have after 7 years, if you invest $1,000.
Ao=1,000 b=2
t=7 k=12
A(t)=1,000(2)^7/12= $1498.31
Rule of 72
72/r%=how long it takes to double.
Ex. How long will it take to double your money at 12% interest?
72/12=6
Chapter 5 is pretty easy, and I should do good on the exam, because I understand most of it due to the fact that it is basically a review of what we've been doing.
Nicala's Post
In class this we learned chapter five which basically the same stuff that we have been doing on the blue part of the pie on aleks. We also started working on the gold and purple parts on the pie which are really hard so I did some of the modules in the green part to make the 14 that i need.
One of things that i worked on aleks is was the lines with slope and a point on the line.
Example problem
A line passes through the point (-5,-3) and has a slope of 2. Write an equation of the line.
The form that you use in this problem is y=mx+b. You put the numbers in place of yr variables.You use -5 as yr x and -3 as yr y.
-3=-5(2)+b
-3=-10+b
you move the -10 to other side of the equal sign and it becomes +1o and you add +10 to -3
which gives me +7
7=b
now you put the slope,2, and yr b,2, back in the problem to get yr answer.
y=2x+7
One of things that i worked on aleks is was the lines with slope and a point on the line.
Example problem
A line passes through the point (-5,-3) and has a slope of 2. Write an equation of the line.
The form that you use in this problem is y=mx+b. You put the numbers in place of yr variables.You use -5 as yr x and -3 as yr y.
-3=-5(2)+b
-3=-10+b
you move the -10 to other side of the equal sign and it becomes +1o and you add +10 to -3
which gives me +7
7=b
now you put the slope,2, and yr b,2, back in the problem to get yr answer.
y=2x+7
Friday, February 25, 2011
Taylor's 13th(I think XD) Blog Review
REVIEW TIME!!!!
This is a review on some of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about arithmetic sequence which is when you add the same number to every term.
This is the formula:
tn= t1 +(n-1)d
d is the number that is being added or subtracted. The n and one by the t are small to.
Ok let's do some examples.
Find the first 5 terms of this equation tn= 8n+4
t1= 8(1)+4 First replace n with 1,2,3, and for and solve the equation.
t2=8(2)+4
t3=8(3)+4
t4=8(4)+4
t1= 12 Then you have your answers.
t2= 20
t3= 28
t4= 36
7,18,29,40.... find the formula for the nth term.
7,18,29,40.... First you need to figure out what is being added or subtracted from
each number. In this case it is adding 11.
tn=t1+(n-1)d Then write out your equation.
tn=7+(n-1)11 Then fill out the parts that you know2.
tn=7+11n-11 Then multiply 5 times n and -1.
tn= -4 + 11n Then you just add up your like terms and this is your answer.
This was the review on some of my notes from Ch. 13.1.
P.s. SOS THERE IS A BEE IN MY HOUSE!!! T-T HELP!! I CAN'T EVEN DO MY HOME WORK!! ALL OF MY STUFF IS IN THE KITCHEN AND IT IS IN THERE TOO!!! T-T
This is a review on some of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about arithmetic sequence which is when you add the same number to every term.
This is the formula:
tn= t1 +(n-1)d
d is the number that is being added or subtracted. The n and one by the t are small to.
Ok let's do some examples.
Find the first 5 terms of this equation tn= 8n+4
t1= 8(1)+4 First replace n with 1,2,3, and for and solve the equation.
t2=8(2)+4
t3=8(3)+4
t4=8(4)+4
t1= 12 Then you have your answers.
t2= 20
t3= 28
t4= 36
7,18,29,40.... find the formula for the nth term.
7,18,29,40.... First you need to figure out what is being added or subtracted from
each number. In this case it is adding 11.
tn=t1+(n-1)d Then write out your equation.
tn=7+(n-1)11 Then fill out the parts that you know2.
tn=7+11n-11 Then multiply 5 times n and -1.
tn= -4 + 11n Then you just add up your like terms and this is your answer.
This was the review on some of my notes from Ch. 13.1.
P.s. SOS THERE IS A BEE IN MY HOUSE!!! T-T HELP!! I CAN'T EVEN DO MY HOME WORK!! ALL OF MY STUFF IS IN THE KITCHEN AND IT IS IN THERE TOO!!! T-T
Sunday, February 20, 2011
Feroz's Blog
This week we learned more on Chapter 12. Woohoo.
One thing I remember clearly is finding the magnitude.
ex. u = <8,3>
Find [u] <-- That's supposed to be an absolute value sign but my laptop doesn't have them.
sqrt(8^2 + 3^2)
= sqrt(73)
And dot product, that's always fun.
ex. <1,3> x <7,4>
* Multiply across and add.
7 + 12 = 19
One thing I remember clearly is finding the magnitude.
ex. u = <8,3>
Find [u] <-- That's supposed to be an absolute value sign but my laptop doesn't have them.
sqrt(8^2 + 3^2)
= sqrt(73)
And dot product, that's always fun.
ex. <1,3> x <7,4>
* Multiply across and add.
7 + 12 = 19
mmmmaaaaarrrrrrryyyyyyyy
This week we learned about chapter 12 and did a billion aleks.
Here are some random notes that i remember because i'm pretty sure i aced that test.
the symbols for absolute value mean the magnitude, which is where you square each term, then take the square root of that number.
l<5,4>l
5^2+4^2=25+16=41
square root of 41 is the magnitude of your vector.
then there's dot product, which they indicate by putting a dot inbetween two vectors. this gives you a number, NOT A VECTOR. hahaha.
<8,0> . <1,2>
then you just multiply across and you may be tempted to believe the numbers multiply across and then stay as a vector, but no, you have to add them.
if the dot product of two vectors equals zero, then they are perpendicular. just thought you'd like to know.
answer 8
they could also ask you to find the cos of the angle between two vectors, but don't fret, its easy if you recognize the examples above.
its the dot product of the vectors over the magnitude of each vector multiplied. you can figure it out by yourselfffffffffff.
Here are some random notes that i remember because i'm pretty sure i aced that test.
the symbols for absolute value mean the magnitude, which is where you square each term, then take the square root of that number.
l<5,4>l
5^2+4^2=25+16=41
square root of 41 is the magnitude of your vector.
then there's dot product, which they indicate by putting a dot inbetween two vectors. this gives you a number, NOT A VECTOR. hahaha.
<8,0> . <1,2>
then you just multiply across and you may be tempted to believe the numbers multiply across and then stay as a vector, but no, you have to add them.
if the dot product of two vectors equals zero, then they are perpendicular. just thought you'd like to know.
answer 8
they could also ask you to find the cos of the angle between two vectors, but don't fret, its easy if you recognize the examples above.
its the dot product of the vectors over the magnitude of each vector multiplied. you can figure it out by yourselfffffffffff.
Dylan's blog-o'-blogs
Ok, so we learned about vectors. We learned about some matrices and other things too, but mostly vectors and stuff. We learned about dot multiplying...which I will be talking about.
Vector 1= (x1,y1), Vector 2= (x2,y2)
To find the dot product:
Vector1*vector2= x1(x2)+y1(y2)
If x=(4,3) & q=(0,6)
then
x*q=4(0)+3(6)
=0+18
=18
(This product will always be a number)
Vector 1= (x1,y1), Vector 2= (x2,y2)
To find the dot product:
Vector1*vector2= x1(x2)+y1(y2)
If x=(4,3) & q=(0,6)
then
x*q=4(0)+3(6)
=0+18
=18
(This product will always be a number)
Nathan's blog
So, we had a test on chapter 12 this past week, and i will just give you the notes and some examples from random sections.
Vectors have a magnitude and a direction.
To add vectors, you add head to tail. The triangle formed by the vectors give the resultant vector.
To add vectors algebraically, you add the components.
Scalar multiplication puts a scalar times each compound.
Parametric equations take only the x portion or y portion.
Dot product
-v1= (x1,y1) v2= (x2,y2)
-v1*v2= x1x2+y1y2
Dot product is a scalar product meaning it gives you a number.
Ex. If u=(3,6), v=(4,2) and w=(-12,-6). Find u*v and v*w. Show that u and v are perpendicular and v & w are parallel.
u*v=(3)(4)+(-6)(2)=0
v*w=(4)(-12)+(2)(-6)=-60
Since the dot product of u*v is 0, then they are perpendicular.
v=(4,2) w=(-12,-6)
v=2(2,1) w=-6(2,1)
V and W are parallel and multiples of each other.
That's it for chapter 12, im glad we're finished with it, because it was kinda hard.
Vectors have a magnitude and a direction.
To add vectors, you add head to tail. The triangle formed by the vectors give the resultant vector.
To add vectors algebraically, you add the components.
Scalar multiplication puts a scalar times each compound.
Parametric equations take only the x portion or y portion.
Dot product
-v1= (x1,y1) v2= (x2,y2)
-v1*v2= x1x2+y1y2
Dot product is a scalar product meaning it gives you a number.
Ex. If u=(3,6), v=(4,2) and w=(-12,-6). Find u*v and v*w. Show that u and v are perpendicular and v & w are parallel.
u*v=(3)(4)+(-6)(2)=0
v*w=(4)(-12)+(2)(-6)=-60
Since the dot product of u*v is 0, then they are perpendicular.
v=(4,2) w=(-12,-6)
v=2(2,1) w=-6(2,1)
V and W are parallel and multiples of each other.
That's it for chapter 12, im glad we're finished with it, because it was kinda hard.
Lawrence's blog
we really didnt do anything new this whole week. it was just review for our chapter test and try to finish the blue chart on the pie. the chapter test wasnt really all that hard to me, it was just alot of finding the equations, dot product and it was just alot of formulas we had to know for chapter 12. i think i got the four by four right and cramers rule i totally forgot. if anyone needs help i can help them with chapter 12 because i got a pretty good hang of it and thank god my friend let me use his laptop to do my blog
Friday, February 18, 2011
Nicala's Blog
The week we have been working on chapter twelve which is vectors and also had matrices like in chapter four but now the matrices are four by four. We also working on the aleks and it gave me an assestment quiz which i bombed and it took like twelve modules from me but its said i went up a couple point i do not know how i went up and fail the assestment. Lately on aleks i have been working on simplify radical expressions which is very so i want to do an example problem of that.
example problem
simplify the following problem
square root of 18x^5y^4z
okay simplify basically means breaking it down to its lowing forms
1. for 18 you can take the square root of nine and that lefts two up the square root
2. x^5 you can take 4 of the Xs because there is five so you put x^2 outside of the square and leave the other x in the square root
3.the same way for y^4 there are four ways so you can take them out and its not an out number so you can take all of it out
4. for the z there is only one of them that why there is no exponent next to it and since there isnt two or more you have leave
it in the square root
the answer is 3x^2y^2 with square root of 2xz
example problem
simplify the following problem
square root of 18x^5y^4z
okay simplify basically means breaking it down to its lowing forms
1. for 18 you can take the square root of nine and that lefts two up the square root
2. x^5 you can take 4 of the Xs because there is five so you put x^2 outside of the square and leave the other x in the square root
3.the same way for y^4 there are four ways so you can take them out and its not an out number so you can take all of it out
4. for the z there is only one of them that why there is no exponent next to it and since there isnt two or more you have leave
it in the square root
the answer is 3x^2y^2 with square root of 2xz
Taylor's 10th(I thinkXD) Blog Review!!
REVIEW TIME!!!
This is a review on the rest of my half of my notes on section 10.1. We will only be doing cos for this section.
The formula that you need to know for this part of the section is this:
cos (alpha+/- beta) = cosalphaxcosbeta -/+ sinalphasinbeta
Alpha and Beta can come from your trig chart. Remember that you need to add or subtract to get the angle that you are looking for.
First Equation
Find the exact value of cos 90 degrees.
cos ( 60+30)= cos 60 degrees x cos 30 degrees - sin 60 degrees x sin 30 degrees First, you need to figure out what angles in the trig chat equal 90 degrees and if you need to subtract or add to get to it. Since 60 degrees plus 30 degrees equals 90 degrees, you will use the subtraction formula for this equation. Then you simply fill out your formula with the information that you were given.
1/2 x square root of 3/2 - square root of 3/2 x ½ Then you find out the trig functions by using your trig chart.
square root of 3/4 - square root of 3/4 Then you multiply the 2 sides by themselves.
cos 120 degrees= 0 Finally you subtract them and you get this as your answer.
That is how you use the formula to find the exact value for cos.
Now lets learn how to simplify.
Simplify cos 45 degrees x cos 60 degrees + sin 45 degrees x sin 60 degrees.
cos 60 degrees x cos 45 degrees + sin 60 degrees x sin 45 degrees First, write out your equation.
cos ( 60 degrees- 45 degrees) Then, since you know that you’re using cos’s formula and subtraction thanks to the formula. You will put it like this.
cos 15 degrees Then you subtract the two degrees and you get this.
cos 15 degrees Since 15 degrees is on the trig chart you will just leave it as the answer to your question.
That is how you use the cos formula in these situations.
This is the review on my notes on the rest of section 10.1.
SO UNTIL NEXT WEEK JA NE!!!
This is a review on the rest of my half of my notes on section 10.1. We will only be doing cos for this section.
The formula that you need to know for this part of the section is this:
cos (alpha+/- beta) = cosalphaxcosbeta -/+ sinalphasinbeta
Alpha and Beta can come from your trig chart. Remember that you need to add or subtract to get the angle that you are looking for.
First Equation
Find the exact value of cos 90 degrees.
cos ( 60+30)= cos 60 degrees x cos 30 degrees - sin 60 degrees x sin 30 degrees First, you need to figure out what angles in the trig chat equal 90 degrees and if you need to subtract or add to get to it. Since 60 degrees plus 30 degrees equals 90 degrees, you will use the subtraction formula for this equation. Then you simply fill out your formula with the information that you were given.
1/2 x square root of 3/2 - square root of 3/2 x ½ Then you find out the trig functions by using your trig chart.
square root of 3/4 - square root of 3/4 Then you multiply the 2 sides by themselves.
cos 120 degrees= 0 Finally you subtract them and you get this as your answer.
That is how you use the formula to find the exact value for cos.
Now lets learn how to simplify.
Simplify cos 45 degrees x cos 60 degrees + sin 45 degrees x sin 60 degrees.
cos 60 degrees x cos 45 degrees + sin 60 degrees x sin 45 degrees First, write out your equation.
cos ( 60 degrees- 45 degrees) Then, since you know that you’re using cos’s formula and subtraction thanks to the formula. You will put it like this.
cos 15 degrees Then you subtract the two degrees and you get this.
cos 15 degrees Since 15 degrees is on the trig chart you will just leave it as the answer to your question.
That is how you use the cos formula in these situations.
This is the review on my notes on the rest of section 10.1.
SO UNTIL NEXT WEEK JA NE!!!
Charlie's.
this week we reviewed for our test.
& yesturday we took our chapter test (it was half mulitple choice & half free response).
i think i got a good grade on it, idk though.
i knew how to do the cramer's rule thing & the 4 by 4 so i think that's the most important thing with the most points..
We also did 14 aleks' modules & have to do 10 modules over the weekend.
(which is a total of 24)
aleks is probably the only reason i'm passing right now.
(i got a 4.0 on my progess report:])
i don't know what else to write on here.
& i have absolutely no idea what example problems to do
because we haven't done new stuff.
K? THANKS. BYE(:
& yesturday we took our chapter test (it was half mulitple choice & half free response).
i think i got a good grade on it, idk though.
i knew how to do the cramer's rule thing & the 4 by 4 so i think that's the most important thing with the most points..
We also did 14 aleks' modules & have to do 10 modules over the weekend.
(which is a total of 24)
aleks is probably the only reason i'm passing right now.
(i got a 4.0 on my progess report:])
i don't know what else to write on here.
& i have absolutely no idea what example problems to do
because we haven't done new stuff.
K? THANKS. BYE(:
Monday, February 14, 2011
Dylan's Blog
I wasn't at school for three days. Blah. I did get the last few notes though.
Determinate's for 3x3 matrices:
1. pick a row r1->most 0s or 1s
2. delete row, delete column
3. repeat for other columns
ex.
|0 3 4|
|2 9 1|
|6 2 1|
this would become
|0 3 4|
|2 9 1|
|6 2 1| x=deleted
this leaves:
|2 9|
|6 2|
Now for the others:
|2 1|
|6 1|
&
|9 1|
|2 1|
Next you take the first number of the row and multiply by the matrix by the number:
|2 9|
|6 2|
|0 3 4|
|2 9 1|
|6 2 1|
4 would be the number used to multiply the matrix
4|2 9|
|6 2|
=
-200 (use ad-bc to find the matrix)
repeat for others, and you should get
0+24-200=-176
Determinate's for 3x3 matrices:
1. pick a row r1->most 0s or 1s
2. delete row, delete column
3. repeat for other columns
ex.
|0 3 4|
|2 9 1|
|6 2 1|
this would become
|0 3 4|
|2 9 1|
|6 2 1| x=deleted
this leaves:
|2 9|
|6 2|
Now for the others:
|2 1|
|6 1|
&
|9 1|
|2 1|
Next you take the first number of the row and multiply by the matrix by the number:
|2 9|
|6 2|
|0 3 4|
|2 9 1|
|6 2 1|
4 would be the number used to multiply the matrix
4|2 9|
|6 2|
=
-200 (use ad-bc to find the matrix)
repeat for others, and you should get
0+24-200=-176
Sunday, February 13, 2011
Feroz's Blog
This week was a blur. I remember finding determinates but that takes a long time to do, especially on a computer. 3x3 might not, but 4x4 just kick my ass. So, since everyone else review things they thought were difficult on Aleks, I'll review something I find difficult as well.
Finding Domain and Range:
Polynomials - Domain (-infinity, infinity)
Range: (-infinity, infinity) if odd
Square Roots - Domain:
1. Set inside equal to 0, 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 0, 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)
Ex. x^3 + 2x - 3
x - 3 = 0
x = 0
Domain: (-infinity, 3) u (3, infinity)
Range: [1, infinity)
Done.
Finding Domain and Range:
Polynomials - Domain (-infinity, infinity)
Range: (-infinity, infinity) if odd
Square Roots - Domain:
1. Set inside equal to 0, 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 0, 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)
Ex. x^3 + 2x - 3
x - 3 = 0
x = 0
Domain: (-infinity, 3) u (3, infinity)
Range: [1, infinity)
Done.
blog...:(
so aleks thinks i'm stupid and wants to know if i can add fractions with common denominators. its SUPER difficult...(sarcasm) so here are some steps and an example for you slow folk. and i actually think i got a negative wrong on the assesment and had to do this module again! its really simple i promise. i would go over what ya'll did friday but i wasn't here and someone should really give me the notes tomorrow.
Problem: (4v-3w)/6v-(9v+5w)/6v
Each fraction has the same denominator, namely
So we subtract the numerators and keep the denominator the same to get
(4v-3w)-(9v+5w)
----------------------
6v
Distributing the negative sign over the parentheses, we have
4v-3w-9v-5w
-----------------
6v
Combining like terms, we get
-5v-8w
---------
6v
This fraction cannot be simplified so that's the answer. have fun.
malorie's blog.
I'm reviewing stuff from the aleks junk because it is the MOST COMPLICATED THING IN WORLD! I get it now though.
Simplifying a ratio of polynomials:
(w^2+6w-7) / (3w^2+33w+84)
To simply this expression you have to factor the numerator and denominator.
*first factor the top out:
w^2+7w-1w-7 = w(w+7)-1(w+7)
=(w-1)(w+7)
*now factor out the bottom:
3w^2+21w+12w+84 = 3w(w+7)+12(w+7)
=(3w+12)(w+7)
=3(w+4)(w+7)
=(w-1)(w+7) / 3(w+4)(w+7)
w+7 can cancel out so your left with
(w-1)/3(w+4)
and that's the final answer.
Simplifying a ratio of polynomials:
(w^2+6w-7) / (3w^2+33w+84)
To simply this expression you have to factor the numerator and denominator.
*first factor the top out:
w^2+7w-1w-7 = w(w+7)-1(w+7)
=(w-1)(w+7)
*now factor out the bottom:
3w^2+21w+12w+84 = 3w(w+7)+12(w+7)
=(3w+12)(w+7)
=3(w+4)(w+7)
=(w-1)(w+7) / 3(w+4)(w+7)
w+7 can cancel out so your left with
(w-1)/3(w+4)
and that's the final answer.
Nathan's blog
I really didn't understand finding the determinants of a 3 X 3 or a 4 X 4. That stuff just blew my mind. I am going to go over the section that is labeled as 3-D.
- absolute value of vector AB= √(x2-x1)^2+(y2-y1)^2+(z2+z1)^2
- midpoint= (x1+x2/2, y1+y2/2, z1+z2/2)
- Equation of a sphere - (x-x0)^2+(y-y0)^2+(z-z0)^2= r^2
- Vector equation - (x,y,z)=(x0,y0,z0)+ t(a,b,c)
- vector addition, magnitude, dot product, all follow same formula with extra variable.
Ex. 1 A sphere has points A(8,-2,3) and B(4,0,7) as endpoints of the diameter.
a. Find the center and radius
b. Find an equation of the sphere.
A.) Center= (8+4/2, 0+-2/2, 7+3/2)
c=6,-1,5
√(4-6)^2+(0-(-1)^2+(7-5)^2= 9
√2^2+1^2+2^2= √9= 3
B.) Equation of the sphere - (x-6)^2+(y-(-1)^2+(z-5)^2=9
So, if you know the formulas for this section it will be pretty easy. I will look over them a few more times before the test, so that I know that I have them memorized. That's all for this blog.
- absolute value of vector AB= √(x2-x1)^2+(y2-y1)^2+(z2+z1)^2
- midpoint= (x1+x2/2, y1+y2/2, z1+z2/2)
- Equation of a sphere - (x-x0)^2+(y-y0)^2+(z-z0)^2= r^2
- Vector equation - (x,y,z)=(x0,y0,z0)+ t(a,b,c)
- vector addition, magnitude, dot product, all follow same formula with extra variable.
Ex. 1 A sphere has points A(8,-2,3) and B(4,0,7) as endpoints of the diameter.
a. Find the center and radius
b. Find an equation of the sphere.
A.) Center= (8+4/2, 0+-2/2, 7+3/2)
c=6,-1,5
√(4-6)^2+(0-(-1)^2+(7-5)^2= 9
√2^2+1^2+2^2= √9= 3
B.) Equation of the sphere - (x-6)^2+(y-(-1)^2+(z-5)^2=9
So, if you know the formulas for this section it will be pretty easy. I will look over them a few more times before the test, so that I know that I have them memorized. That's all for this blog.
Friday, February 11, 2011
Taylor's 9th (I Think XD)blog Review!!
REVIEW TIME!!!
This is a review on my notes from lesson 8.1.
In this lesson we will be solving for THETA. In order for you to be able to solve for Theta you need to get the trig. function by itself and then you need to take an inverse of it. An inverse will have two answers, but there may be exceptions to this rule if the question asks you to just find one quadrant. Any way, you need to find where an angle is based on the trig function and then you need to tell if it is negative or positive.
These are the steps:
First, you need to take the inverse of the number so that you can find the First Quadrant angle.
Then you simply need to remember this in order to find the other answers:
Quadrant 2, you need to make the first quadrant angle a negative and add 180 degrees to it.
Quadrant 3, you just need to add 180 degrees to the first quadrant angle.
Quadrant 4, you need to make the first quadrant angle a negative and add 360 degrees to it.
Now let's solve some problems!!
(REMEMBER I DO NOT KNOW HOW TO PUT A GRAPH ON HERE SO PLEASE NOTE THIS. WHEN I SAY DRAW A QUARDNENT PLANE DO SO AND CHECK THE ONES THAT I TELL YOU TO OK!!)
Solve for Sin A= 2/5 for A.
Sin A = 2/5 First, write the equation down.
A= Sin^-1(2/5) Then put the equation in inverse form like I just did.
A= 23.578 degrees Then you find the inverse and make sure to round to the 3rd place after the decimal. Ok, now draw you quadrant plane. Now since our Sin is positive we need to put a check in the quadrants where Sin or Y would be positive. Which is the First and Second Quadrant. So since we have the First Quadrant already we just need to find the Second Quadrant.
-23.578 degrees + 180 degrees= 156.422 degrees Finally you just use the formula that I gave you to find the angle of the Second Quadrant and this is your answer.
So your answers should be 156.422 degrees and 23.578 degrees.
Let's try cosine next.
Solve for Cos H = 7/9 for H.
Cos H = 7/9 First, write the equation down.
H= Cos^-1(7/9) Then put the equation in inverse form like I just did.
H= 38.942 degrees Then you find the inverse and make sure to round to the 3rd place after the decimal. Ok, now draw you quadrant plane. Now since our Cos is positive we need to put a check in the quadrants where Cos or X would be positive. Which is the First and Fourth Quadrant. So since we have the First Quadrant already we just need to find the Fourth Quadrant.
-38.942 degrees + 360 degrees= 321.058 degrees Finally you just use the formula that I gave you to find the angle of the Fourth Quadrant and this is your answer.
So your answers should be 38.942 degrees and 311.81 degrees .
That is the review on my notes on Lesson 8.1.
This is a review on my notes from lesson 8.1.
In this lesson we will be solving for THETA. In order for you to be able to solve for Theta you need to get the trig. function by itself and then you need to take an inverse of it. An inverse will have two answers, but there may be exceptions to this rule if the question asks you to just find one quadrant. Any way, you need to find where an angle is based on the trig function and then you need to tell if it is negative or positive.
These are the steps:
First, you need to take the inverse of the number so that you can find the First Quadrant angle.
Then you simply need to remember this in order to find the other answers:
Quadrant 2, you need to make the first quadrant angle a negative and add 180 degrees to it.
Quadrant 3, you just need to add 180 degrees to the first quadrant angle.
Quadrant 4, you need to make the first quadrant angle a negative and add 360 degrees to it.
Now let's solve some problems!!
(REMEMBER I DO NOT KNOW HOW TO PUT A GRAPH ON HERE SO PLEASE NOTE THIS. WHEN I SAY DRAW A QUARDNENT PLANE DO SO AND CHECK THE ONES THAT I TELL YOU TO OK!!)
Solve for Sin A= 2/5 for A.
Sin A = 2/5 First, write the equation down.
A= Sin^-1(2/5) Then put the equation in inverse form like I just did.
A= 23.578 degrees Then you find the inverse and make sure to round to the 3rd place after the decimal. Ok, now draw you quadrant plane. Now since our Sin is positive we need to put a check in the quadrants where Sin or Y would be positive. Which is the First and Second Quadrant. So since we have the First Quadrant already we just need to find the Second Quadrant.
-23.578 degrees + 180 degrees= 156.422 degrees Finally you just use the formula that I gave you to find the angle of the Second Quadrant and this is your answer.
So your answers should be 156.422 degrees and 23.578 degrees.
Let's try cosine next.
Solve for Cos H = 7/9 for H.
Cos H = 7/9 First, write the equation down.
H= Cos^-1(7/9) Then put the equation in inverse form like I just did.
H= 38.942 degrees Then you find the inverse and make sure to round to the 3rd place after the decimal. Ok, now draw you quadrant plane. Now since our Cos is positive we need to put a check in the quadrants where Cos or X would be positive. Which is the First and Fourth Quadrant. So since we have the First Quadrant already we just need to find the Fourth Quadrant.
-38.942 degrees + 360 degrees= 321.058 degrees Finally you just use the formula that I gave you to find the angle of the Fourth Quadrant and this is your answer.
So your answers should be 38.942 degrees and 311.81 degrees .
That is the review on my notes on Lesson 8.1.
Charlie.
This week we did Aleks & we have been doing stuff with vectors.
vectors have a magnitude and a directions.
to add vectors you add head to the tail.
the triangle formed by the vectors give the resultant vector.
to add vectors algebraically you add components.
to find the magnitude > square root of x^2 + y^2
magnitude is done a|^->|
scalar multiplication puts a scalar times each component.
geometric representation = picture.
head ----tail
.________>
Example:
.__> A
.
|
|
|
|
\/ B
~~A+B =
.__>
|
|
|
|
\/
~~2A + 1/2B
.__.__>
|
|
\/
vectors have a magnitude and a directions.
to add vectors you add head to the tail.
the triangle formed by the vectors give the resultant vector.
to add vectors algebraically you add components.
to find the magnitude > square root of x^2 + y^2
magnitude is done a|^->|
scalar multiplication puts a scalar times each component.
geometric representation = picture.
head ----tail
.________>
Example:
.__> A
.
|
|
|
|
\/ B
~~A+B =
.__>
|
|
|
|
\/
~~2A + 1/2B
.__.__>
|
|
\/
Sunday, February 6, 2011
Feroz's Blog
Another long day doing Aleks stuff. Anyway, Go Packers.
I went to school 2 out of 5 days last week, so forgive me if I'm not sure what I'm talking about.
No one really went over 12-1, so I don't what that's all about. Something about lines. I don't know. So I guess that leaves Matrices. And I'm going to use 1x2's because anything bigger just looks ridiculous.
ex.
H = [2 3]
G = [4 5]
Find H + G
H + G = [6 8]
ex. 2 Find H x G
You can't really multiply them because they're dimensions don't work. The "inside" numbers have to match. ex. 2x3 3x4
*On a side note, Division requires you to multiply the matrix by the other's inverse. Again, that still requires the dimension's "inside" numbers to match.
I went to school 2 out of 5 days last week, so forgive me if I'm not sure what I'm talking about.
No one really went over 12-1, so I don't what that's all about. Something about lines. I don't know. So I guess that leaves Matrices. And I'm going to use 1x2's because anything bigger just looks ridiculous.
ex.
H = [2 3]
G = [4 5]
Find H + G
H + G = [6 8]
ex. 2 Find H x G
You can't really multiply them because they're dimensions don't work. The "inside" numbers have to match. ex. 2x3 3x4
*On a side note, Division requires you to multiply the matrix by the other's inverse. Again, that still requires the dimension's "inside" numbers to match.
Nathan's blog
Okay, so section 12-1 involved drawing little lines and connecting them. That's about it for that section. So, since that can't get me 150 words, I will just review matrices again. You can add, subtract, or multiply matrices but you can't divide them.
Adding Matrices:
[4 5] + [-9 2] = [-5 7]
[7 8] [5 -3] [12 5]
Subtracting Matrices:
[7 18] - [5 4] = [2 12]
[9 4] [-8 2] [17 2]
Multiplying Matrices:
[2 5] X [3 10] = [11 30]
[7 6] [1 2] [27 82]
That's only some of the things on matrices. You also have scalar multiplication, Cramer's rule, and finding the inverse and determinant of a matrix.
Cramer's rule is:
Dx/d
Dy/d
Scalar multiplication is when you solve without using Cramer's rule.
The determinant of a matrix is ad-bc.
The inverse of a matrix is 1/ the determinant of the matrix.
That's all I got on matrices. Next time I will probably review something different.
Adding Matrices:
[4 5] + [-9 2] = [-5 7]
[7 8] [5 -3] [12 5]
Subtracting Matrices:
[7 18] - [5 4] = [2 12]
[9 4] [-8 2] [17 2]
Multiplying Matrices:
[2 5] X [3 10] = [11 30]
[7 6] [1 2] [27 82]
That's only some of the things on matrices. You also have scalar multiplication, Cramer's rule, and finding the inverse and determinant of a matrix.
Cramer's rule is:
Dx/d
Dy/d
Scalar multiplication is when you solve without using Cramer's rule.
The determinant of a matrix is ad-bc.
The inverse of a matrix is 1/ the determinant of the matrix.
That's all I got on matrices. Next time I will probably review something different.
BLOOGGGGG
This week, we learned about matrices, which was reallllyyrealllyy realllyy easy whereas what i'm attemping to learn in aleks is driving me crazy. i mess up once and then fix it because it'll be an arithmetic error or i would have just typed it wrong on the last on before its "done" but then it says "oh, do it two more times". AHHHHHHHHHHHHHH. i'm gonna show you some of this maddness.
The thing i was stuck on was a number +/- a fraction OVER another number +/- another fraction! its so confusing!
but i don't have any problems to show you! i can make one up and show you the procedure, but i'm not gonna work it all the way through because its not gonna be pretty.
5-6x/2x^2
-------------
2+3/8x
say this is your problem, you have to find a commom denominator of the fractions inside the fraction. which would be 8x^2
then you multiply everything by this common denominator!
you would get:
45x^2-48x^3/2x^2
-------------------
16x^2+8x^2/8x
then you can reduce and get:
45x^2-24x
-----------
16x^2+x
then usually on aleks, it can be factored and reduced to a simple fraction, but i'm not gonna try mine, have fun once all ya'll get to that part of the aleks.:)
The thing i was stuck on was a number +/- a fraction OVER another number +/- another fraction! its so confusing!
but i don't have any problems to show you! i can make one up and show you the procedure, but i'm not gonna work it all the way through because its not gonna be pretty.
5-6x/2x^2
-------------
2+3/8x
say this is your problem, you have to find a commom denominator of the fractions inside the fraction. which would be 8x^2
then you multiply everything by this common denominator!
you would get:
45x^2-48x^3/2x^2
-------------------
16x^2+8x^2/8x
then you can reduce and get:
45x^2-24x
-----------
16x^2+x
then usually on aleks, it can be factored and reduced to a simple fraction, but i'm not gonna try mine, have fun once all ya'll get to that part of the aleks.:)
Kaitlyn's blogg
This week we did a test on matrices and we learned 12-1. We also worked on aleks a lot, which is good because its pretty easy and will help bring up everyone's grades in that class. I forgot how to do 12-1 and i forgot to bring my binder home, so i will go over matrices :)
Adding and subtracting matrices is pretty straightforward. you just add or subtract the number across from it in each matrix.
Ex:
[2 3] + [7 5] = [9 8]
[1 5] [1 0] [2 5]
When you multiply matrices, you must make sure the dimensions are correct. if you have a 3x2 matrix and a 2x1 matrix, you can multiply this because the two middle numbers (2 and 2) are the same.
Ex:
[2 3] x [7 5]= [14+ 3 10+0]= [17 10]
[1 5] [1 0] [7+5 5+0] [12 5]
Adding and subtracting matrices is pretty straightforward. you just add or subtract the number across from it in each matrix.
Ex:
[2 3] + [7 5] = [9 8]
[1 5] [1 0] [2 5]
When you multiply matrices, you must make sure the dimensions are correct. if you have a 3x2 matrix and a 2x1 matrix, you can multiply this because the two middle numbers (2 and 2) are the same.
Ex:
[2 3] x [7 5]= [14+ 3 10+0]= [17 10]
[1 5] [1 0] [7+5 5+0] [12 5]
lawrence's blog
there really isnt much to say about what we did this week. this week all we did was take a test on matrices and went over chapter 12-1. we also spent alot of time on aleks which is good because it will bring up alot of our grades. for those who didnt understand matrices it was pretty easy.
imma show yall if a matrix is able to be multiplied. for example:
3x2 2x2 you can multiply these because the last and first number are the same and you will have a 3x2
4x8 9x7 you cant multiply these because the first and last number arent the same
now imma add one just to be nice :)
2 5 7 + 4 5 6 = 6 10 13
3 9 1 6 7 8 9 16 9
that is all for now and if you need help with anything just let me know. i can help
imma show yall if a matrix is able to be multiplied. for example:
3x2 2x2 you can multiply these because the last and first number are the same and you will have a 3x2
4x8 9x7 you cant multiply these because the first and last number arent the same
now imma add one just to be nice :)
2 5 7 + 4 5 6 = 6 10 13
3 9 1 6 7 8 9 16 9
that is all for now and if you need help with anything just let me know. i can help
Friday, February 4, 2011
Taylor's Blog Review
REVIEW TIME!!!
These are the review notes on what we learned in section 7.5
You will need to know the following so that you can work out the equations.
Tan Theta=y/x
Cot Theta = x/y
Sec Theta = r/x
Csc Theta = r/y
Ok now that we know what Tan, Cot, Sec, and Csc are lets use them to solve the following equations.
Since Sin Theta = 2/8 0<> 90 degrees find the meaning of the other five trig functions.
(REMEMBER I DON’T KNOW HOW TO PUT A GRAPH ON THE COMPUTER SO I WILL JUST TYPE IT OUT.)
Ok, we know the since Sin is positive then the y needs to be positive and that can only happen on the second and first quadrant. Also since 2/8 is supposed to be less than 90 degrees we need to be in the first quadrant. Ok now we need to find x. You can draw a triangle and put in the info that you know to help you.
A^2+2^2=8^2
We will use the Pythagorean theorem to help us find x.
A^2+4=64
a^2=60
We will then use the exponents to get theses answers and then we will subtract 4 from 64 and get 60 then we will find the square root of a and 60 and the answer will be the square root of 60.
Now we just fill in the blanks.
Cos x/r= square root of 60/8
Tan y/x= (since we can’t have a square root at the bottom we will multiply it by the square root of 60 and reduce and get this) square root of 60/30
Cot x/y= square root of 60/2
Sec r/x=(since we can’t have a square root at the bottom we will multiply it by the square root of 60 and reduce and get this) 2 times the square root of 60/15
Csc r/y = 8/2 (reduced is) 4
Now lets try to find the reference angle of tan.
Find the reference angle for Tan 43 degrees.
First we find what quadrant its in. Which is the first because of the degrees which means tan is positive.
Tan 43 degrees
Since 65 degrees is already in-between 0 and 90 degrees we don’t need to subtract and since it’s not on the Trig chart that means that Tan 43 degrees is your answer.
That is 7.5 review Until next time!
These are the review notes on what we learned in section 7.5
You will need to know the following so that you can work out the equations.
Tan Theta=y/x
Cot Theta = x/y
Sec Theta = r/x
Csc Theta = r/y
Ok now that we know what Tan, Cot, Sec, and Csc are lets use them to solve the following equations.
Since Sin Theta = 2/8 0<> 90 degrees find the meaning of the other five trig functions.
(REMEMBER I DON’T KNOW HOW TO PUT A GRAPH ON THE COMPUTER SO I WILL JUST TYPE IT OUT.)
Ok, we know the since Sin is positive then the y needs to be positive and that can only happen on the second and first quadrant. Also since 2/8 is supposed to be less than 90 degrees we need to be in the first quadrant. Ok now we need to find x. You can draw a triangle and put in the info that you know to help you.
A^2+2^2=8^2
We will use the Pythagorean theorem to help us find x.
A^2+4=64
a^2=60
We will then use the exponents to get theses answers and then we will subtract 4 from 64 and get 60 then we will find the square root of a and 60 and the answer will be the square root of 60.
Now we just fill in the blanks.
Cos x/r= square root of 60/8
Tan y/x= (since we can’t have a square root at the bottom we will multiply it by the square root of 60 and reduce and get this) square root of 60/30
Cot x/y= square root of 60/2
Sec r/x=(since we can’t have a square root at the bottom we will multiply it by the square root of 60 and reduce and get this) 2 times the square root of 60/15
Csc r/y = 8/2 (reduced is) 4
Now lets try to find the reference angle of tan.
Find the reference angle for Tan 43 degrees.
First we find what quadrant its in. Which is the first because of the degrees which means tan is positive.
Tan 43 degrees
Since 65 degrees is already in-between 0 and 90 degrees we don’t need to subtract and since it’s not on the Trig chart that means that Tan 43 degrees is your answer.
That is 7.5 review Until next time!
Charlie's.
This week we reviewed & took a test on matrices [Chapter 14].
& like everyday we did aleks.com thingy.
(i still have like a wholeee bunch left to 'master')
& we just started chapter 12 yesterday.
chapter 12 is something weird.
i honestly already fergot how to do it.
the only thing i know is that you draw things,
& then if it says to add them, you connect the two arrows.
& on the two arrows thingy, there's a head and a tail.
so, since i totally fergot how to do it ALREADY.
i'll just do an example of matrices.
EXAMPLE:
[ 2 8 1 ] [ 8 2 7 ]
[ 6 5 7 ] + [ 4 6 9 ] =
[ 8 1 3 ] [ 4 6 2 ]
[ 10 10 8 ]
[ 10 11 16 ]
[ 12 7 5 ]
& like everyday we did aleks.com thingy.
(i still have like a wholeee bunch left to 'master')
& we just started chapter 12 yesterday.
chapter 12 is something weird.
i honestly already fergot how to do it.
the only thing i know is that you draw things,
& then if it says to add them, you connect the two arrows.
& on the two arrows thingy, there's a head and a tail.
so, since i totally fergot how to do it ALREADY.
i'll just do an example of matrices.
EXAMPLE:
[ 2 8 1 ] [ 8 2 7 ]
[ 6 5 7 ] + [ 4 6 9 ] =
[ 8 1 3 ] [ 4 6 2 ]
[ 10 10 8 ]
[ 10 11 16 ]
[ 12 7 5 ]
Monday, January 31, 2011
Feroz's Blog
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.
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.
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.
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