Okay, so this week might have been the easiest week ever except for monday. I thought the exam was tough, but i guess i just didn't study enough and i'll just do better next time. Though for the rest of the week we did powerpoints and learned some easy work. I like when we do the projects because it gives us a break from using our brains haha. But this week we started Chapter 7. Chapter 7 is starting off simple, but you have to memorize your formulas. We learned about converting radians to degrees and degrees to radians.
EXAMPLE: Convert to radians.
1.) 225 degrees
225 X pi/180
225/180 pi
45/36 pi
= 5/4 pi
2.) 30 degrees
30 X pi/180
30/180 pi
= 1/6 pi
Convert to degrees.
1.) 3/4 pi
3/4 pi X 180/pi
the pi cancels out
540/4
= 135 degrees
2.) pi/6
pi/6 X 180/pi
the pi cancels out
180/6
= 30 degrees
We also learned how to find measures of central angles, find approximate diameters, etc. I sorta understand, but can someone explain it to me again. It's kinda confusing. Thanks!!! :)
Saturday, October 17, 2009
Sunday, October 11, 2009
reflet. 8
This week was very simple and easy all we did was review everything for our exam
one thing i really understood was longs and expressing y interm of x
ln y =1/3ln(x) +ln 4
ln y =4x^1/3
so basically your solving for y and that problem is half way done for you...
expanding and simplfying logs
log m- 4log n
when to logs are subtracting it would show up as division when simplifying
log m/log n^4
one thing i really understood was longs and expressing y interm of x
ln y =1/3ln(x) +ln 4
ln y =4x^1/3
so basically your solving for y and that problem is half way done for you...
expanding and simplfying logs
log m- 4log n
when to logs are subtracting it would show up as division when simplifying
log m/log n^4
reflection 8
One thing I really understand how to do from the study guide is this thing (i forget what its called)
2x^4 + 4x^2 + 2 = 0
g=x^2
so the equation becomes: 2g^2 + 4g + 2 = 0
now simply factor it: (2g^2 + 2g) (2g + 2)
2g(g + 1) 2(g+1)
g=-1 and g=1
now plug in to:
x^2 = -1 and x^2 = 1
so your roots will be: (1,0) (-1,0) (i,0) (-i,0)
------------------------------------------------------------------------------------
I forgot how to do the p/q thing i could use some help.
2x^4 + 4x^2 + 2 = 0
g=x^2
so the equation becomes: 2g^2 + 4g + 2 = 0
now simply factor it: (2g^2 + 2g) (2g + 2)
2g(g + 1) 2(g+1)
g=-1 and g=1
now plug in to:
x^2 = -1 and x^2 = 1
so your roots will be: (1,0) (-1,0) (i,0) (-i,0)
------------------------------------------------------------------------------------
I forgot how to do the p/q thing i could use some help.
relection 8
ok so we did our study guides this week and presented our power points, so from the study guides i am relfecting on something that i totally didnt understand until today, haha
i^1=.25=i
i^2=.50=-1
i^3=.75=-i
i^4=whole #=1
Ex: i^16= whole # so it is 1
chapter 6 is on the test right...wish me luck guys on this exam cuz i need it really bad!
i^1=.25=i
i^2=.50=-1
i^3=.75=-i
i^4=whole #=1
Ex: i^16= whole # so it is 1
chapter 6 is on the test right...wish me luck guys on this exam cuz i need it really bad!
Reflection #8
Well, this week was...stressful. But the study guide wasn't that hard. I weren't many things I had questions about and the things I didn't understand, I already figured out.
The only thing I don't completely understand is:
Ch. 4 #10 (free response)
I understand why it's even...but that's about it. I don't see how it's symmetric, I have no idea what a/b or y-axis means...????? any help??
One thing I do understand....
The word problems:
"I am building a rectangular pen to keep all my pigs from escaping the farm. I sue the side of my outhouse as part of the enclosure and only have 100ft of wood to use. What is the mas. area I can enclose for my pigs?"
1. draw figure
2. the width of either side is x and the length is 100-x/2
*A=length times width
3. 100=2x x=50
4. 100-(50)/2 =25
5. 25 x 50
6. 1250ft^2
SUPER EASY!! ^^
The only thing I don't completely understand is:
Ch. 4 #10 (free response)
I understand why it's even...but that's about it. I don't see how it's symmetric, I have no idea what a/b or y-axis means...????? any help??
One thing I do understand....
The word problems:
"I am building a rectangular pen to keep all my pigs from escaping the farm. I sue the side of my outhouse as part of the enclosure and only have 100ft of wood to use. What is the mas. area I can enclose for my pigs?"
1. draw figure
2. the width of either side is x and the length is 100-x/2
*A=length times width
3. 100=2x x=50
4. 100-(50)/2 =25
5. 25 x 50
6. 1250ft^2
SUPER EASY!! ^^
Reflection #8
Parabolas:
to find axis of symmetry:
x=-b/2a
Vertex:
a. (-b/2a, f(-b/2a)
b. complete the square to get into vertex form-
y=(x+a)^2+b ((A & B ARE #s))
(-a,-b)
Focus:
1/4p= coefficient of x^2 then add p.
Directrix is p units behind vertex
always x= or y=
so subtract p.
EXAMPLE:
x^2+1
-V:(0,1)
-focus: 1/4p=1
p=1/4
-directrix: y=1-1/4
y=3/4
to find axis of symmetry:
x=-b/2a
Vertex:
a. (-b/2a, f(-b/2a)
b. complete the square to get into vertex form-
y=(x+a)^2+b ((A & B ARE #s))
(-a,-b)
Focus:
1/4p= coefficient of x^2 then add p.
Directrix is p units behind vertex
always x= or y=
so subtract p.
EXAMPLE:
x^2+1
-V:(0,1)
-focus: 1/4p=1
p=1/4
-directrix: y=1-1/4
y=3/4
I'm having a little trouble with some of the scratch work for chapter 4 test..
Refelction #8
Well i'm not gonna sit here and B.S. about all kinda of crap my brain is fried and i'm super tired. Ill say this if u havent been studyin or doin ya study guides good luck gettin a decent F. All this week i did was study guides, study guides, and more study guides. I dreamed anout math, i thought about math, my food taste like math, i just need a break from it all. And guess what when i did hang out with people or was talking to sumone guess what they brought up, MATH!!!!!!!!!!!! Well, i hope i do well 2morrow looking forward to gettin a B for the first 9 weeks. Remember mrs. Robinson it is my first year here i would hate to start off with anything lower that a B. I'll be ya best friend!!!!!!!!!!!!!!!!!!!!!
Reflection #8!
So this week was quite fun, study guides, powerpoints, etc!
Although we didn't really learn many new things, we did learn parabolas:
In order to find the axis of symmetry:
x=-b/2a
Vertex:
a. (-b/2a, f(-b/2a)
b. complete the square to get into vertex form-
y=(x+a)^2+b ((A & B ARE #s))
(-a,-b)
Focus:
1/4p= coefficient of x^2 then add p.
*if opens up, add to y-value from vertex.
*if down, subtract y-value from vertex.
*if opens right, add to x-value from vertex.
*if opens left, subtract.
Directrix is p units behind vertex
always x= or y=
so subtract p.
EXAMPLE:
x^2+1
-V:(0,1)
-focus: 1/4p=1
p=1/4
-directrix: y=1-1/4
y=3/4
---------now for the things i don't know----------
i'm having difficulty finding the correct scratchwork for number 9 on chapter 3.
Solve for x:
2x^4+14x+12
the answer given by B-rob was (i,o)(-i,o)(i(square root of 6)) (i(-square root of 6))
how? when worked, i got x=2,-3, -1, --(i couldn't find my fourth term)
i must be mixing something up?
Oh, and is Chapter 6 on the exam? i think so but idk. i'll just study it anyway!
thanks! :)
Although we didn't really learn many new things, we did learn parabolas:
In order to find the axis of symmetry:
x=-b/2a
Vertex:
a. (-b/2a, f(-b/2a)
b. complete the square to get into vertex form-
y=(x+a)^2+b ((A & B ARE #s))
(-a,-b)
Focus:
1/4p= coefficient of x^2 then add p.
*if opens up, add to y-value from vertex.
*if down, subtract y-value from vertex.
*if opens right, add to x-value from vertex.
*if opens left, subtract.
Directrix is p units behind vertex
always x= or y=
so subtract p.
EXAMPLE:
x^2+1
-V:(0,1)
-focus: 1/4p=1
p=1/4
-directrix: y=1-1/4
y=3/4
---------now for the things i don't know----------
i'm having difficulty finding the correct scratchwork for number 9 on chapter 3.
Solve for x:
2x^4+14x+12
the answer given by B-rob was (i,o)(-i,o)(i(square root of 6)) (i(-square root of 6))
how? when worked, i got x=2,-3, -1, --(i couldn't find my fourth term)
i must be mixing something up?
Oh, and is Chapter 6 on the exam? i think so but idk. i'll just study it anyway!
thanks! :)
Reflection 8
This week we worked on study guides from all our old tests for our exam tomorrow. In which I probally will fail, but besides that we also did our powerpoint presentations that I did not get to actually do yet.
When looking at the study guides I realized I did not know how to do was # 6 on chapter 4 find the inverse of y= square root of x-1.
And to freshen up my memory ima do the i chart.
i^1=.25=i
i^2=.50=-1
i^3=.75=-i
i^4=whole #=1
Ex: i^84= whole # so it is 1
When looking at the study guides I realized I did not know how to do was # 6 on chapter 4 find the inverse of y= square root of x-1.
And to freshen up my memory ima do the i chart.
i^1=.25=i
i^2=.50=-1
i^3=.75=-i
i^4=whole #=1
Ex: i^84= whole # so it is 1
Reflection #8
For this week we learned how to tell the shape of a graph if not in standard form and how to graph parabolas. The thing I understood the most is how to know the shape of a graph if not in standard form and here is an example:
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
-you use the formula B^2-4AC
1. if it is –ve then it is a circle or an ellipse.
2. if B^2-4AC=0 then it is a parabola
3. if B^2-4AC is a +ve then it is a hyperbola.
Example: 4x^2+6xy+2y^2+8x+3y+5=0
4^2-4(4)(2)
=-16
**This would be a circle or an ellipse.
Now for something I didn’t quite understand is how you find the directrix, the focus, and how to graph parabolas. I don’t understand the p thing either.
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
-you use the formula B^2-4AC
1. if it is –ve then it is a circle or an ellipse.
2. if B^2-4AC=0 then it is a parabola
3. if B^2-4AC is a +ve then it is a hyperbola.
Example: 4x^2+6xy+2y^2+8x+3y+5=0
4^2-4(4)(2)
=-16
**This would be a circle or an ellipse.
Now for something I didn’t quite understand is how you find the directrix, the focus, and how to graph parabolas. I don’t understand the p thing either.
reflection 8
An entire week to work on study guides for the exam..not bad. this was a very productive week for me, and I should understand everything i dont get. the only thing new thing we learned was parabolas, and they seem to be easy to do
The study guide really did help me, and i have a fresh memory of chapters 1 thru 5
The study guide really did help me, and i have a fresh memory of chapters 1 thru 5
reflection 8
yeah this week was awesome. working on power points and doing study guides, thats whats up. i think i pretty much understand everything thats on the study guides. and mrs robinson said that if you know how to do every problem on the study guides that you will do good on the exam, so hopefully i do good, ha. now just because i said that ima fail..yay. one thing that we did ectually take notes on and learned this week was parabolas.
EXAMPLE:
x^2+1
vertex: x=-b/2a= 0/2(1) = 0
(0,1)
focus: 1/4p= 1/1
p=1/4
(0,1+1/4) = (0,5/4)
Directrix: y=1-1/4
y=3/4
and your graph is going to have a u on the top of the x=axis.
EXAMPLE:
x^2+1
vertex: x=-b/2a= 0/2(1) = 0
(0,1)
focus: 1/4p= 1/1
p=1/4
(0,1+1/4) = (0,5/4)
Directrix: y=1-1/4
y=3/4
and your graph is going to have a u on the top of the x=axis.
Reflection 5 redo
Well, this week in advanced math we pretty much did study guides all week. We worked on study guides so we really didnt learn too much new and that gave us time to catch up and stuff. But i think that the easiest thing to do out of everything we had to do was find the vertex. To find vertex you use -b/2a. Then plug your answer back into the problem that you had to start off with.
Like: 4x^4+8x
a)Plug into vertex form.
-8/2(4)=-1
b)Plug into original equation.
4(-1)^4+8(-1)=-2
c)Put in point form.
(-2,0)
I get mostly everything we did I just need to learn how to get better at all of that graphing stuff. It takes me a while to remember how do to do that while im working so i need to learn that better before the exam.
Like: 4x^4+8x
a)Plug into vertex form.
-8/2(4)=-1
b)Plug into original equation.
4(-1)^4+8(-1)=-2
c)Put in point form.
(-2,0)
I get mostly everything we did I just need to learn how to get better at all of that graphing stuff. It takes me a while to remember how do to do that while im working so i need to learn that better before the exam.
REFLECTION #8
So another week has gone by. Actually, it went by pretty fast if you ask me. The things we've been learning in advanced math have been pretty easy to grasp, but I do have to say that Chapter 6 was probably the hardest chapter so far. I'm deffinitely not going to say Chapter 6 was easy because I got very confused on some things with parabolas and hyperbolas. Anyway, Monday we didn't learn the usual amount of information that we normally do. We just learned an easy way to find out what shape an equation is---whether it's an ellipse, a circle, a hyperbola, or a parabola. Then on Tuesday we learned about parabolas. We were taught how to find the vertex, focus and directrix of those types of problems. Then, Wednesday, Thursday, and Friday, we were given time in class to work on our study guides which helped a lot considering we have quite a few of them.
Alright, so I thought the easiest thing that we learned this week was determining what shape an equation is.---whether it's an ellipse, circle, hyperbola, or a parabola. All you have to do is use on simple formula. The formula is b^2 - 4ac. **If you notice, this formula also exists in the quadratic formula and it is also the formula used to find the discriminate. (or to find how many x intercepts you have when graphing a parabola)
So here's an example:
*You are given the equation and you are asked to figure out what the shape of it is. This is the equation: 2x^2 + 5xy - 3y^2 + 6x + 4y + 12
*Remember, all you have to do is use b^2 - 4ac. If the number you get is negative, then the shape is either a circle or an ellipse. If it's positive, then it's a hyperbola. If it equals zero (0) then it is a parabola.
So you get:
b^2 - 4ac
(5)^2 - 4(2)(-3)
25 - 8(-3)
25 + 24 = 49 >> which is positive, so it is a hyperbola
***Now for what I didn't quite grasp this week. Something that I had a little trouble with was something like "Find the equation of an ellipse.." when you are only given a vertex and a focus. These types of problems stump me so if anyone has an easy way to explain them more to me that would be appreciated :)
Overall, this week wasn't ALL that bad.
Alright, so I thought the easiest thing that we learned this week was determining what shape an equation is.---whether it's an ellipse, circle, hyperbola, or a parabola. All you have to do is use on simple formula. The formula is b^2 - 4ac. **If you notice, this formula also exists in the quadratic formula and it is also the formula used to find the discriminate. (or to find how many x intercepts you have when graphing a parabola)
So here's an example:
*You are given the equation and you are asked to figure out what the shape of it is. This is the equation: 2x^2 + 5xy - 3y^2 + 6x + 4y + 12
*Remember, all you have to do is use b^2 - 4ac. If the number you get is negative, then the shape is either a circle or an ellipse. If it's positive, then it's a hyperbola. If it equals zero (0) then it is a parabola.
So you get:
b^2 - 4ac
(5)^2 - 4(2)(-3)
25 - 8(-3)
25 + 24 = 49 >> which is positive, so it is a hyperbola
***Now for what I didn't quite grasp this week. Something that I had a little trouble with was something like "Find the equation of an ellipse.." when you are only given a vertex and a focus. These types of problems stump me so if anyone has an easy way to explain them more to me that would be appreciated :)
Overall, this week wasn't ALL that bad.
Reflection 8
sooo..I think this week was my favorite in advance math. We worked on study guides which gave us a break from learning something new, and also gave us time to understand the things we didn't understand. Some of us also gave our presentations (or tried to)..ha. Anyways, On all the study guides the easiest thing is finding vertex. To find vertex you use vertex form which is
-b/2a. Then plug your answer back into the original equation. Put answer into point form.
Example: 2x^2+4x
*Plug into vertex form.
-4/2(2)=-1
*Plug into original equation.
2(-1)^2+4(-1)=-2
*Put in point form.
(-2,0)
--------------------------------------
I pretty much know how to do everything, I just have to remember how to do it. And remember all the formulas.
-b/2a. Then plug your answer back into the original equation. Put answer into point form.
Example: 2x^2+4x
*Plug into vertex form.
-4/2(2)=-1
*Plug into original equation.
2(-1)^2+4(-1)=-2
*Put in point form.
(-2,0)
--------------------------------------
I pretty much know how to do everything, I just have to remember how to do it. And remember all the formulas.
Reflection #8
This was the best week so far. All we did in class was work on our study guides, do our powerpoint presentations, and learn hyperbolas and parabolas. This week went by really fast though. We also did our chapter 6 test. I really don't know what to blog about becuase we didn't really go into a chapter and learn different things. But i did understand parabolas.
_______________________________________________________________
EX:
1) You have to find the axis of symmetry. x=-b/2a
2) Find the vertex. (-b/2a, f(-b/2a)) or complete the square to get vertex form.
y= (x+a)^2 +-b (a and b are numbers) (-a,b)=vertex
3) Find the focus: 1/4p= the coefficient of x^2 and then add p
**If opens up, add to y value from vertex, if opens down, subtract
**If opens right, add to x value to vertex, if opens left, subtract
4) Directrix is p units begind vertex, so subtract p.
**If oopens up, subtract: if opens down, add to y value of vertex
**If opens right, subtract: If opens left, add x value.
________________________________________________________
EX: x^2 + 1
Vertex 0/2= 0
Focus 1/4p = 1/1 p=1/4
(0, 1+1/4)>>>>(0, 5/4)
Directrix y=1-1/4>>>>>y=3/4
Graph:
_________________________________________________________
I can do the parabola thing when i am looking at my notebook, but when i get to a test, i freeze. So if anyone has any tricks or things that help you remember, or even if you just give me another example, i might remember it better. Thankss.
_______________________________________________________________
EX:
1) You have to find the axis of symmetry. x=-b/2a
2) Find the vertex. (-b/2a, f(-b/2a)) or complete the square to get vertex form.
y= (x+a)^2 +-b (a and b are numbers) (-a,b)=vertex
3) Find the focus: 1/4p= the coefficient of x^2 and then add p
**If opens up, add to y value from vertex, if opens down, subtract
**If opens right, add to x value to vertex, if opens left, subtract
4) Directrix is p units begind vertex, so subtract p.
**If oopens up, subtract: if opens down, add to y value of vertex
**If opens right, subtract: If opens left, add x value.
________________________________________________________
EX: x^2 + 1
Vertex 0/2= 0
Focus 1/4p = 1/1 p=1/4
(0, 1+1/4)>>>>(0, 5/4)
Directrix y=1-1/4>>>>>y=3/4
Graph:
_________________________________________________________
I can do the parabola thing when i am looking at my notebook, but when i get to a test, i freeze. So if anyone has any tricks or things that help you remember, or even if you just give me another example, i might remember it better. Thankss.
Reflection #8
Okay, this week was pretty much overall good. The only thing we really learned new was things to do with parabolas, so I'll explain that. I like exam weeks because we usually get a break from doing homework, and we can just review what we already know, or for some people, should've learned.
For parabolas,
Axis of symmetry: x = -b/2a
There are 2 ways to find the vertex:
(-b/2a, f(-b/2a))
OR
complete the square to get vertex form
y = (x+a)² + b
a & b are #'s
(-a,b) = vertex
Focus:
1/4p = coefficient of x²
then add p
Directrix: is p units behind the vertex
Always x= or y=
So subtract p
*If opens up, subtract; if opens down, add from y-value of vertex.
*If opens right, subtract x-value
*If opens left, add x-value
Example: x² + 1
Vertex: x=-b/2a = 0/2(1) = 0
(0,1)
Focus: 1/4p = 1
4p = 1
p = 1/4
(0, 1 + 1/4)= (0, 5/4)
Directrix: y = 1 - 1/4
y = 3/4
Well, overall, I'm glad that this week was kind of like a rest week, and all we learned was about parabolas. So, just to get things clear, the focus is as you would say, inside the vertex, and the directrix is outside, essentially?
Reflection 8
I liked this week a whole lot better than any of the other weeks. The reason is mostly because we had a take home chapter 6 test, and we just reviewed. We did learn some more information on parabolas, but that was an easy concept to grasp. For parabolas you find axis of symmetry, vertex, focus, and directrix.
Example: x^2+1
1.) vertex: x= -b/2a = 0/2(1) = 0 0^2+1=1
(0,1)
2.) focus: 1/4p=1/1= p=1/4
(0,1+p) = (0,1+1/4)
= (0,5/4)
3.) directrix: y=1-1/4
y=3/4
Overall, finding parabolas are pretty simple. Since it is exam week and we have to review chapters 1-6, we have to rework all our test for study guides and it gives us extra points :)
So if anyone wants to help and remind me about inverses from chapter 4, that would be great!
Example: x^2+1
1.) vertex: x= -b/2a = 0/2(1) = 0 0^2+1=1
(0,1)
2.) focus: 1/4p=1/1= p=1/4
(0,1+p) = (0,1+1/4)
= (0,5/4)
3.) directrix: y=1-1/4
y=3/4
Overall, finding parabolas are pretty simple. Since it is exam week and we have to review chapters 1-6, we have to rework all our test for study guides and it gives us extra points :)
So if anyone wants to help and remind me about inverses from chapter 4, that would be great!
Subscribe to:
Posts (Atom)