blog #somethin

so im a little late but i been sick all weekend. last week we learned about derivatives and the many forms a question about derivatives might take. so the formula for the derivative of a tangent line is f(x)= f(x+deltax)-f(x)/deltax. say u had to find the derivative of x^3+3x:

the first thing u would do is plug in ur formula: lim over deltax -> 0 (x+deltax)^3 + 3(x+deltax) - (x^3+2x)/deltax

(x+deltax) (x^2+3xdeltax+deltax^2)

x^3 + 3x^2deltax + xdeltax^2 + x^3deltax + 3x^2deltax^2 + xdeltax^3 / deltax

which comes to: x^3 + 4x^2 + 4xdeltax + deltax^2

then u plug in 0 for deltax: x^3 + 4x^2

tada

Final Reflection

we learned a wide variety of things this year in advanced math. the year went by really fast, and i did have trouble on some things. trig was possibly the hardest thing this year, and precalculus is where we continue from next year. those who are in calc next year, get ready for more math videos. this class wasnt extremely scary as i thought it would be, i mean at least we got past 7 chapters, unlike chemistry. hopefully calculus wont be as crazy and confusing as this year

FINAL REFLECTION #2

Mkay, so my last Final Reflection seemed to be the wrong format. :x

So i'll just talk about my year in this one, instead of explaining something we should already know. :)


K, so I definitely wish I could say "i SO could have done better", but really, this year I tried my hardest and still didn't do as well as I'd hoped. Unfortunately, math club or not, Math is my worst subject. I can learn something in class, retain it for 55 minutes, but when i'm alone and having to comprehend what I learned that day, I'm S.O.L. Everything with numbers and variables seems to go in through one ear and out of the other. What's the most frustrating part about it all is that I LEARN IT, but then... I forget... EVERYTHING. Sorry for the hour of complaints and whining, just thought i'd let this out on my last blog. For the optimistic and HAPPY things about advanced math, I really did learn quite a few things that I will take with me next year in Calculus. Advanced math really helped me on ACT too! I had fun in the class when I understood something, stressed when I didn't, and triumphed when I finally learned. Advanced math is filled with many mixed emotions, assignments, blogs, and EDEE. Too bad we lost our only senior, and too bad he was the one to make the class interesting! On the bright side, I think we learned a lot more when he was gone. :p

I hope next year I can take what I've retained, written, and studied with me to calculus and succeed as much as I'd hoped for THIS year. I'll try harder next year, and hopefully over the summer, my summer packet will help me to remember everything and refresh my memory for a fresh start in calc!

I hope this blog, and all of the blogs I re-did or sent in late can help me go from a C to a B in this class! :)

All in all, i enjoyed my year with you guys!


THANKS B-ROB!
Can't wait for calc!





HAPPPPPY SUMMMMMMMER<3

Final Reflection

We learned sooo many things in advanced math this year. I really didn't think that we could learn so much in just one year. Of course there were things that i had no clue how to solve or i didn't understand at all. But I eventually asked a friend or re-read my notes and then I understood better. In the middle of the year i had to get a tutor:/, which was a wastee!, because this guy had no clue what he was doing and didn't know what we were learning because his class hasn't gotten to it yet, so i was screwed. The only thing I knew how to do was the problems with the trig chart because i made myself memorize that thing because I knew i was going to need it with pretty much every problem. There were also a lot of things that I understood very quickly and they were easy for me. The tests in this class were pretty hard, but I mean if you studied they were probably very easy, ha. But as of right now, my notebook went missing:( The last thing i remember doing with it is giving it to Jammi to copy some notes, and she said she doesn't have it. Everythinggg is in there, all the tests that need to be turned in and my notes, study guide..everything. I don't know what to do. Im about to cry. But anyway, this class wasn't as bad as i thought it was going to be, but I am definately nott taking calculus next year! I can barely hang in advanced math, there is no way im going to be able to keep up with the people in calculus. BYEEE!:)

FINAL REFLECTION

Yes the year is finally coming to an end. It's been a very good year and i am very proud of all of the things i learned in adv math this year. I learned so much and ms robinson has been by far the absolute best math teacher i have ever had. Even though her work can sometimes be a pain i will be always greatful for the things she taught me and the stuff we did in her class. She forced us to learna and retain the information. I never had to do so many chapter test and keep redoing them. The bridge project was super fun too, boy how i love to stay up all night and make bridges haha. One thing I found out in advance math is you have to learn your formula's and there are plenty of them to learn. If you dont know your formulas it is hard to even try to work a formula. I always think subjects are going to be hard at the beginning of the year, but it turned out not to be so bad. A few of the things we learned at the beginning of the year were easy, then it kept building up from there. I'm glad I got to know some of the concepts that we have learned, and I enjoyed doing other activities that helped me understand things better. Im glad we did so much work with the trig and the trig chart. We finished 14 chapters of math and begin working on calculus stuff. I know im ready and prepared for next year. O, we also read a book that was super easy and helped with me in english as well. Well this has been definitly a year to remember, i wouldnt have gotten stuff like this at lutcher. Thanks for everything and i shall be seeing all of yall next year. I'm still preparing for our final exam, and i hope i do well. Happy summer to everyone!

Final Reflection

We learned a lot of things this year in Advanced Math. I always think subjects are going to be hard at the beginning of the year, but it turned out not to be so bad. A few of the things we learned at the beginning of the year were easy, then it kept building up from there. I'm glad I got to know some of the concepts that we have learned, and I enjoyed doing other activities that helped me understand things better. The bridge project demonstrated good trigonometry usage and it was fun doing. Also, reading Flatland was also "fun", being that we kinda had a break from doing a lot of math. I actually might have learned a life lesson through this book though -.- but, overall, this class was pretty easy, sometimes, and sometimes it was difficult. But, I'll say one thing for sure, this year FLEW. Where'd the time go? Oh well. Well, I still don't understand limits completely, so I guess I'll have to study up some more, and hopefully pass this final exam Tuesday. At least it's multiple choice and has some easy beginning chapter material on it. Okay, I'm done now. HAPPY SUMMER.

final ONE!!

well we learned a lot of stuff this year, like 14 chapters of stuff to be exact. from having a formula for everything to reading a book about dimensions to having to memorizing the TRIG CHART!!. i think everyone had a successful year with everything we learned this year. some things were easier than others and some were really hard to get the concept. like logs and triangles were some easy things we learned this year. and a thing i never caught on to was the trig identies. overall i think everyone learned a lot that will be useful in college.

This year in advance math we learned a lot of stuff. One thing I found out in advance math is you have to learn your formula's and there are plenty of them to learn. If you dont know your formulas it is hard to even try to work a formula. Some of the easy things we did was logs. Codensing and working logs seemed to be easy. Also if you dont know your trig chart and unit circle you will probably lose a bunch of points and not know how to do everything in the trig part of the class. Everything revolves around the trig chart. Another thing that I thought was going to be really stupid was the book we had to read about the square in a flatland, but it had a lot of inside meaning and related to real life concepts. Overall it wasnt that bad of a class, but I needa make a good grade on my exam so I don't fail this last nine weeks.

FINAL REFLECTION!

So this year has been a long year and i thought advanced would be super hard (and it kind of was), but was not too bad, just took a lot of studying. But it also took a lot of memorizing and studying, we learned everything from factoring to trigonometry to the beginning of calculus. Well for this last blog i'll give examples of stuff i understood well throughout the year.

EXAMPLES:
Condensing.
1.)logm + log7 + 4logn
= log7mn^4
2.)5loga + logd + log6
= log6da^5
3.)logn - 3logh -logy
= n/yh^3
4.)4logt - logc
= t^4/c

Expanding.
1.)log5gh^2
= log5 + 2logh +logg
2.)m^3b^7/f
= 3logm + 7logb – logf

This is about trig review. I had to study hard but i guess i ended up doing decent with trig. This geometry type stuff just isn't my thing. I did understand a good bit though.
EXAMPLE:
C = 90 degrees
b = 7
c = 12
First i found angle A
cos A = 7/12
A = cos^-1(7/12)
= 54.315 degrees
Second i found angle B
sin B = 7/12
B = sin^-1(7/12)
= 35.685 degrees
Next i found length of a
tan 54.315 = a/7
(7) tan 54.315 = a/7 (7)
a = 7 tan 54.315
a = 9.747
Finally i found the area
A = 1/2 bh
= 1/2 (7) (9.747)
= 34.115

I'm gonna review something we learned at the very begining of trig, SOHCAHTOA.
sin=opposite leg/hypotenuse
cos=adjacent leg/hypotenuse
tan=opposite leg/adjacent leg
*You use SOCAHTOA for right triangles

We learned a lot of formulas we have to memorize like: sum and differences with sin and cos, half and double angles, and many more. I found the easiest formula to work with was tan alpha + tan beta/1-tan alpha tan beta. I also found the sin(alpha + beta) was easy.

We learned how to find the angle of inclination, which i found was really easy compared to some stuff we learn. We also learned about amplitudes, periods, vertical shifts, etc. There are some formulas we had to learn to be able to work these problems:
1.) For a line
m=tan alpha where m=slope and alpha=angle of inclination
2.) For a conic
tan 2 alpha=B/A-C
3.) For a conic if A=C then
a=pi/4

EXAMPLE:
Find the angle of inclination.
2x+5y=15
m=-2/5 tan alpha=-2/5 Checks are in the II and IV area and 21.801 degrees in I
alpha=tan^-1(-2/5)
180-21.801 alpha ~ 158.199 degrees, 338.199 degrees
158.199+180

I also understood how to do domain and range very well. The types of the domain and range problems i understood the most were the fractions and polynomials. (oo => stands for infiniti)
EXAMPLE: y=x^3+4x^2+12
For any type of polynomial the domain would be (-oo,oo) and for the range, odd:
(-oo,oo); quadratics: [vertex,oo] or [-oo,vertex].
The answer to this problem would be Domain: (-oo,oo) Range (-oo,oo)
EXAMPLE: 5x+4/x^2-4
First, set the bottom of the fraction equal to zero.
x^2-4=0
+4+4
x^2=4
x=+ or - 2
Your answer then comes to,
Domain: (-oo,-2)u(-2,2)u(2,oo)

I'll give an example of one of the problems from our Chapter 1 test:
13.)Solve for x by completing the square x^2 - 4x = 9
First you must dived b by 2, then square the answer of b divided by 2 then plug it in to the eqn.
-4/2 = (-2)^2 = 4 x^2-4x+4=9+4 Second you take the -2 from parenthesis and square it minus x, also must add 9+4 on the other side of equal sign. (x-2)^2=13 Third, you take the square root.
square root of (x-2)^2= square root of 13. Next you add 2 to each side giving you: x=2+ 0r - square root of 13. and Finally put the answer in point form: (2+squareroot of 13, 0) (2-squareroot of 13, 0).

The final thing we learned was stuff dealing with calculus.
Examples:
1. lim (4x^2+3) = 4(2)^2+3 = 19
x>2
2. lim x^2+5x-6/x-1 = (x+6)(x-1)/x-1 = (1+6) = 7
x>1

Basically i did decent in this class, and we started learning calculus so i hope do good in that next year, and this is our LAST BLOG :)

Final Reflection

Alrightttt, since I did my last reflection totally wrong I have to redo it. I really need to learn how to pay attention hah. Anyways, this year was a very hard year for me but I still some how managed to pull out with B's. At the beginning of the year I thought I'd never make it, but through out the year it got easier and easier as I started to catch on to how B-rob taught. I learned a lot this year, and actually for the first time still remember how to do a lot of it. I realized advance math is all about formulas, and if you don't study your formulas and notes you will never pass the class. I learned that each chapter builds off of the beginning chapters so it's smart to understand the beginning lessons before getting into the other chapters. One of the chapters you really had to pay attention on was identites, we pretty much used identies in everything for trig. One of the things I really remember how to do is logs and and finding if something is symetric (those are really random things to remember..I know). But I think I remember those the best because there the easiest. I also will never forget the trig chart after drilling it in my head, haha. Another thing that helped me was blogs, commenting on everyones blog helped me remember the lesson we learned in class that week, and reassured me if I was doing something wrong. That's pretty much all I have to say about this year, i'm not really sure what else to put. Overall, this year has been my hardest year yet in math..and I hope next year is easier.

FINAL REFLECTION!! :D

I'm soooo happy to say the time has finally come for our final reflection of the year! (: And I can honestly say I've learned a lot this year. Not only concepts that were new to me, but just that practicing and studying ALWAYS helps and makes a difference in this class. I remember in the beginning of the year I was complaining about how hard everything was and how the blogs were so hard and how everything was stressful. But now that I look back on the entire year, it really wasn't bad at all. Once I got used to how things were, I knew what to expect and I always made sure I asked my questions so I completely understood everything before I took a test. And I know that everytime we were assigned homework, I always thought it was too much, but I'm glad I did it because that deffinitely helped me understand things better. Commenting blogs during the week also was beneficial because if someone didn't know how to do something and I did, I was glad that I was catching on and understood what I was doing. Also, no matter how annoying the Chapter tests were (that we worked through about a hundred times for each chapter hah) that helped a lot, because everytime we had to retake the tests for a study guide, it helped me remember things better and I didn't have to look back at my notes that much. I can honestly say that I remember basically everything we learned this year, because the information was pretty much glued to my brain. Especiaaallllyyy the trig chart! I know that thing like I know my prayers haha. And identities is something that I thought was reaaaally easy this year. As long as you knew them, and knew how to apply them to a problem, you were good. OHH! And another thing you should be pround about Ms Robinson!! (: >> From all the things we went over this year I was able to help out my friends a lot in Algebra 2 with some things they were learning. I felt special because I thought it was extremely easy haha but anyway I was glad I could help them out. SOOOOO anyway, overall I don't think this year was really hard, but I can honestly say this is the most I've learned and can actually remember in a math class. All the other years I'd learn concepts, but by the time we went off for the holidays I'd forget so I had to relearn them. I don't think that's gonna happen this time. So I'm glad about that and I'm sooo ready for Calculus next yearrrr! (: ..not to sound like a nerd or anything ha

Final Reflection

So this is my last reflection of the year! I'm kind of excited about that, but i'm not too excited that i will be doing them next year. Yes, i am taking calculus, but i wasn't too happy about it. I'm starting to realize that it isn't going to be as bad as everyone is making it out to be. Its just math, what can be so hard about it? I mean we are learning pre-calc right now, and it is pretty simple. Well, if yo pay attention and do your homework, then you shouldn't have a problem with calculus. But we're going to have a total of like five people in that class next year. But anyway, i've learned A LOT of stuff in advanced math this year. In the beginning of the year, i was kind of nervous about the work and stuff, but i got used to it. I think that everyone in our class was so nervous because the advanced math class before us, was making it seem SUPER hard. But it really was not that bad. So i think that calculus is going to be easy next year. But i can definitley say that i have learned to factor. That is the one out of many things that i will take with me out of that class. I can also say that i, comfortably know the trig chart, and all of its meanings. I have also learned how to solve logs, how to apply formulas, and how to solve triangles. (and yes ashton i do know what a triangle is). Oh, and the bridge project! That was the funnest project ever, except for the fact that my group had to do it all in one night, because we don't pay attention to directions! Oh well, it was still fun because we had to pull an all nighterrr (: it was super duper fun! Not to mention the fact that we won still, our group's bridge still held the most weight, and the glue was still wet. But i have to say, that this year's math class, i have learned WAYYY more than i have ever learned in math. I will definitley take most, if not all, of the things that i've learned with me to college, and in everyday life. But guys, guess what?! We are almost out of school, soooo, FINISH STRONG (: we only have one week left. And everyoone should take the optional test on fridayyy, because it would be an extra grade. Plus this chapter is superrr easy!

Last Reflection

This is my last reflection of the year. And I was remembering one of the easy things we did which is logs. This will probally show up on the final exam.

Examples of some logs:

log 7 of 49 = 2 log x of 36 = 2
7^2 =49 x^2 = 36
x = 6

log 3 of 27 = 3 log x of 64 = 3
3^3 =27 x^3 = 64
x = 4

log 2 of 32 = 5 log x of 100 = 2
2^5 =32 x^2 = 100
x = 10

Logs should be easy if you just look over them.

Last Reflection

For my final reflection…
I’ve learned a lot this year in Advanced Math. We learned some stuff that just branched off of Algebra II at the beginning of the year, such as the rational root therom, domain and range, shapes of graphs, word problems, and reviewing basic factoring. But besides all of this, we learned a LOT of trig. (This also happens to be my least favorite part of this year.☺) First, the basic with SOHCAHTOA and the six trig functions. There were lots of triangles and angles, and most of it was very confusing (not to mention the bridge project *shudders*). Then there was the trig chart, one thing I actually know by heart! And now, as the year is ending, we begin Calculus. It’s surprisingly easy, but I am NOT looking forward to taking Calc AB next year. But anyway, all in all, I learned a lot this year.

School's almost out!!

5/16

This week went by uber fast!! The intro to calculus we learned was pretty easy. i actually understand it. Then we did worksheets, practice problems, and whatnot. hopefully, this calculus stuff is on the exam!! then maybe i'll do semi-good on it. but trig wise... mehhhh... i gotta go back and study for that stuff. and formulas. and the trig chart. i sure hope this week goes by as fast as last week did... so i can get this year over with and relax in the summertime

LAST ONE!!!!!!!!!!!!

Last Reflection

Formulas:Cos(α +/- β)=cos α cos β -/+ sin α sin βsin(α +/- β)=sin α cos β -/+ cos α sin βsin x + sin y= 2 sin x + y/2 cos x-y/2sin x - sin y= 2 cos x + y/2 sin x-y/2cos x + cos y= 2 cos x + y/2 cos x-y/2cos x - cos y= 2 sin x + y/2 sin x-y/2

tan (α + β)=tan α + tan β/1-tan α tan βtan (α - β)=tan α - tan β/1+tan α tan β

sin2α=2sin α cos αcos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1tan 2α = 2tan α /1-tan^2 αsin α/2= +/- √1-cos α/2cos α/2= +/- √1+ cos α/2tan α/2= +/- √1-cos α or 1 + cos α=sin α/1+cos α=1-cos α/sin α

something that i understood the most was section 2 here's some examples:
tan α = 2 and tan β=1
find tan (α - β)
= tan α + tan β/1-tan α tan β
=2+1/1-(2)(6)
=3/-1
=-3

Find the exact value of: tan 15+tan 30/1-tan 15 tan 30
tan α = 2 and tan β=1
find tan (α - β)
= tan (15 + 30)
=tan (45)
=1

im gonna miss you guys, especially you b-rob!! im gonna miss failing every test this summer, but dont worry, i'll fail em all next year too

last blog

so we are now finished with the advanced math book, and now we're moving on to calculus which is surprisingly easy without a catch

Continuity, which means to draw a graph without picking up the pencil, is basically what chapter 1 is about

there are 4 types of discontinuity:

1. vertical asymptote (can't touch this)
2. removable (hole)
3. jump (enough said)
4. Osciallations (not dealing with it)

when you see f(#)it means youre only looking for the colored in y value

example:

limit as x -> 2^+ means: the limit as x approaches 2 from the right

by looking at a certain graph with several curves, lines, and points, you should be able to find the types of discontinuities

Finite Limits
y-values that the GRAPH is approaching
they do not exist if there are 2 different y values

functions are also used in calculus

f(x) = -x^2 + 4x

find lim f(x) x->3

-(3)^2 + 4(3)
-9 + 12 = 3

also used with trig functions

f(x) = sin x
lim f(x) x->pi/4

sin 45 = sqrt 2 /2

composite functions (f[g[x]]) are also used

Reflection 5/16

For my final reflection, I will talk about, well I forget the name of it, but the f(x) an g(x) thing....

ok

f(x)=x+4 and g(x)=x-2

what is f(4)
f(4)= (4) + 4= 8

what is g(3)
g(3)= (3) - 2 = 1

what is f(g(2))

g(2)=0, so f(0) = 4

see simple, you just have to break it down.

what is f(2g(5))

g(5)= 3, so multiply it by 2 because its 2g(5) and you get 6. So plug that into f....
f(6)= 10

See its simple. You just have to break it down one step at a time. Don't look at the problem as a whole or you will get confused. Jus look at it one step at a time. Like look at g(5) first...solve that, the 2g, then plug that answer into f(x) and you'll be fine. Just remember, ONE STEP AT A TIME!!!

Final Reflection

Just some stuff:

You use completing the square to solve a quadratic equation when factoring doesn’t work. This can only work when 1 is the coefficient of x².

For example:

x² + 6x - 2 = 0

x² + 6x = 2

x² + 6x + 9 = -2 + 9

(x + 3)² = 7

x + 3 = √7

x = -3 ± √7

(-3 + √7,0) (-3 -√7,0)

Final Reflection Stuff:

The most important thing we learned this year was probably learning the Trig Chart. Everything we did always came back to knowing the trig chart. An activity that help me this year was: All Seniors Take Calc to help determine were sin cos and tan are positive on a graph. All goes in the first quadrant, all of them are positive in quadrant one, Seniors goes in quad two because sin is positive there, Take goes in quad three because Tan is positive there, and Calc goes in quad four because thats where its positive.

REFLECTION 5/16

Timeeee for another bloggggg. OH BOYYY. Anyyyway, this week we started learning calculus, and it's pretty easy so far. I'd say the main thing we learned this week was limits (..but I don't have my notebook in front of me so I'm not sure haha) Well I'll give a few examples of finding limits without using the chart or calculator.

Ex. 1) lim x^2+2x+1/x^2-1
x>3
*when you're asked to find the limit, the first thing you want to do is plug in the number you're given for x into the equation. Start plugging in with the bottom of a fraction (if you have a fraction) because if you get zero as the denominator, then you have to use the chart.
*So for this problem, you get 8 as the denominator so you're okay.
*Also, to make things easier, you can factor the problem before you plug in.
*So you get (x+1)(x+1)/(x+1)(x-1)
*And you can cancel out the (x+1)'s and you get (x+1)/(x-1)
*Now you can plug in 3 for x. And you should get:
4/2...which equals 2

Ex. 2) lim 4x^3 + 6
x>1
*Since this problem isn't a fraction you don't have to worry about getting zero as a denominator. So you know you can just plug in
*So you get 4(1)^3 + 6
*And that equals 10

Ex. 3) f(x)=2x+3 g(x)=4x^2+1 *Find lim f(g(x))
x>2
*Alright this is just like a composite function except all you have to do is add a limit in
*So when you set it up you get: f(lim 4x^2+1)
x>2
*Now all you have to do is plug 2 into that equation
*And you get 4(2)^2+1
= 17
*So now you have f(17)
*So all you have to do with that is plug 17 into the "f" function you're given in the beginning of the problem
*So you get: 2(17)+3
*And that gives you 37

**Allrrigghtttyyy that's about all the examples I can show, because we also did alot of things with graphs so I can't show that on here.
(:

reflection may 16

This week went by pretty fast actually, i thought. The calculus that we learned at the beginning of the week is pretty easy. I really think i understand it:) Then we did worksheets, practice problems, and little quizes in class that i thought i did well on. So if we have a good bit of this new stuff on the exam, i think i will do pretttty good on itt. But other than this new stuff, i will have to study and look over the old stuff that we learned, like doman and range, ugh im so bad at those. And those problems where you have to memorize the formulas, yeah i suck at those too. Im very proud of myself with the trig chart though:), i studied that thing sooo much for the trig exam, all i will have to do is look over it a few times and i'll have it back again in my head. So, hopefully that this coming week will fly by so that we will only have exams left, which are half dayss! Then the summmerrr:)

Reflection.

Alright so we finished advance math and started calculus, which was suprisingly easy :) hopefully it just stays that way. I'll just explain some things that we need to know for the exam.

**Functions!
f(x) and f(g(x))

Okay, for functions you'll always get one, two, or more equations>>f(x)>> that you will use to solve individual problems like this for example:

Example 1

f(x)=x+4 g(x)=2x-3
(these are the two equations that you are given^)

a.) find (f + g) (x) >>By the plus sign, this notation tells you that you're going to add your two equations together like this:

x + 4 + 2x - 3
= 3x - 1

b.) f(g(x)) >>This is a composite function. What this function is telling you to do is plug in the "g" equation every time you see an x in the "f" equation. like this:

(2x-3) + 4
(*you always want to put what you plugged in in parenthesis in case you have to distribute a negative.)

then you get >> 2x-3+4
= 2x + 1


c.) find g(6) >> this notation has a number in parenthesis. Whenever you see a number in parenthesis you know that your answer should be a number.

So the function g(6) tells you to plug in the number 6 every time you see an x in the "g" equation.
so you would get >> 2(6) - 3 = 9

**Domain & Range!

There are four different ways you can find the domain and range depending on the equation.

The first one is domain & range of all polynomials.
The domain of all polynomials is (-00,00)
The range of all odds: (-00,00)
The range of quadratics: [vertex,00)/positive[-00,vertex]/negative

The second one is the domain and range of fractions.
To find the domain and range of fractions you just set the bottom=0. After doing that solve for y. Then set up your intervals. Which would be (-00,closest number to -00) (<--closest number to -00,closest numberto 00) (closest number to 00,00)

The third one is domain and range of absolute value.
Absolute value for domain is (-00,00)
Range if positive (shift,00)
Range if negative (00,shift)
*Shift is the number that's outside the absolute value.

The fourth one is the domain and range of square roots.
First you would set whats inside the square root equal to 0. Then set up a number line. Try values on either side, (to do this you plug in numbers that you find on the left andthe right of the number on your number line into the equation inside the square root)Eliminate anything negative. Then set upintervals.

Here are some things to remember:
**Notice all domain is (-00,00)
**When talking about vertex in polynomials you use vertex form.
**Domain is x values, Range is y.
**() means included - used for +/-infinity
**[] means not included - used on numbers

4th and final reflection.

So, i'll take this time to complain and rejoice in the SAME blog.
SHOUT OUT TO MALERIE BULOT WHO REMINDED ME ABOUT MY FINAL BLOG TODAY, a little late, YET BETTER THEN NOTHING!

Wow, ADVANCED MATH IS OVER. ON TO THE CALCCCC! I might fail miserably, but ya can't say i won't try!
:)

and for the second night in a row, i am still at edee's working on this spoof. we stayed here til 2am yesterday, and i have a feeling we're on the same roll this time.

i'm going to dieeeee, :/


here's my last blog!


To start off, there are two types of inequalities, regular inequalities and absolute value inequaties. In absolute value inequalities there are two different types determined by the signs if it is not an equal sign. They can either be "and" or "or" inequalities. "And" inequalities always have a less than symbol (<) and "or" inequalities always have a greater than symbol (>). Also, (something important to remember) in absolute value inequalities you always get two answers. I'm sure everyone remembers how to work these problems but here's a few examples:

Ex. 1.) 4x + 1 > 13
*you would just solve this like a regular equation even though it does not have an equal sign. *and for this problem you would only get one answer because it does not have an absolute value symbol.
*so you subtract 1 from 13 and you get 4x > 12
*divide by 4 and you get that x > 3

Ex. 2.) 2y - 4 = 12 (*assume there is an absolute value thing around 2y-4)
*the first thing you want to do with this problem is set up two equations. since in absolute value you know you will get two answers
*your first equation will be the original equation but without the absolute value symbol like this:
2y - 4 = 12
*and your second equation will be the same equation except you change the 12 to -12 like this:
2y - 4 = -12
*then you solve both equations for y and you get that y = -4 and 8

Ex. 3.) 3x - 4 + 5 < 27 (*assume there is an absolute value thing around only 3x - 4)
*the first thing you have to do is isolate the absolute value and to do that you have to get rid of the 5. so you subtract the 5 over from the 27 and your new equation is:
3x - 4 < 22
*the next thing you notice is that this equation has a < sign and that means it is an "and" equation so you have to set it up a certain way like this:
-22 < 3x - 4 < 22
*then you solve the equation first by adding 4 over to all sides then you get this:
-18 < 3x < 26
*then you divide by 3 on all sides
your final answer is: -6 < x < 26/3

Next comes slope! First of all, there are 3 different equations of slope. There's slope intercept form, point slope form, and standard form.
Slope intercept is y = mx + b.
Point slope form is y - y1 = m (x - x1).
standard form is Ax + By = C


Here are a few examples of problems:

Ex. 4.) Find the slope of the two points (4,1) (3,0)
*to solve this you use the formula for slope which is: m = y2 - y1/x2 - x1
*so you get...0 - 1/3 - 4 and that gives you -1/-1 which equals 1

Ex. 5.) Find the equation of the line 3x + 4y = 12 that is perpendicular to the point (3,2).
(*by the way, I'm not sure if I worded that problem right but hopefully you should know what I'm talking about right?)
*so the first thing you have to do is find the slope of the equation you are given. 3x+4y=12
*first you subtract 3x over and you get 4y = -3x + 12
*then you divide by 4. and you get that y = -3/4x + 3
*So your perpendicular slope of that equation is 4/3. (because you take the negative reciprocal of the orginal slope of the equation)
*Then to put it in an equation including the point you are given, you use the point slope formula.
*So your final answer is y - 2 = 4/3 (x - 3)

reference angles and exact answers. *Remember that reference angles can only be between 0 and 90 degrees. Here are some examples:

Ex. 6.) Find the reference angle of sin 236. (*assuming that there's a degree sign after 236)
*the first thing you have to do is figure out what quadrant 236 is in. it's in the 3rd quadrant.
*sin relates to the y axis and in the third quadrant, the y axis is negative so for the reference angle, sin has to be negative
*and to find the reference angle of 236 you have to subtract 180. and you get 56 degrees.
*so your reference angle is equal to -sin 56. and since this is not on the trig chart, to find the exact answer you would have to type it into your calculator.

3rd late reflection.

Notice, by this time I'm practically DEAD.
this blog thing is pretty tiring after a while,
and PLUS i'm doing last minute corrections to our mu alpha theta spoof. :(

On the bright side, functions were pretty simple to me so i should probably explain them:

First, let's start with the fact that if you find the function of x, or f(x), you will find y.

Adding or subtracting functions:
(f+g)(x)
f(x)+g(x)

(f-g)(x)
f(x)-g(x)

example:

f(x)=x+1
g(x)=x-4

(f+g)(x)= x+1+x-4
(f+g)(x)=2x-3

Multiplying functions:
(fxg)(x)
f(x) x g(x)

f(x)=2x+4
g(x)=4x-6

(2x+4)(4x-6)
F.O.I.L.= (8x^2-12x+16x-24)
=8x^2+4x-24

Finding a function of another function:
(fog)(x)=
f(g(x))

f(x)= 2x+4
g(x)= x-4

plug in the g(x) equation into the f(x) equation where ALL x's appear.

2(x-4)+4
distribute.
2x-8+4
(fog)(x)=2x-12

Second Late reflection.

As I stated in the last reflection,
i'm DEFINITELY behind.

So here's another blog!

I'm just explaining some of the stuff we learned in chapter 13 that I actually thought I understood.

* to find a term
arithmetic- tn = t1 + (n-1)d
where n=term #, t1=first term, d=what you add, tn=term you're looking for
geometric- tn = t1 x r^(n-1)
where r=what you multiply by
* to find the sum
arithmetic- sn = n(t1 + tn) / 2
geometric- sn = t1(1 - r^n) / 1-r
tn-1 = previous term
tn-2 = previous to the previous term

**Arithmetic sequences are when you have to add or subtract to get the next term. Geometric sequences are when you have to multiply to get the next term.

Examples:Find the formula for this arithmetic sequence:
3, 5, 7,...
tn = 3 + (n-1)2
tn = 3 + 2n - 2
tn = 1 + 2n

*These are easy you just plug in the previous term to get your answer.
You add 2 to each number to get the next number so plug 2 in for (n-1)
Solve, then you'll get the answer.

Find the next four numbers in the sequence:
t1 = 1 & tn = 3(tn-1) - 11, 2, 5, 14, 41,...

*For these you just keep plugging in the previous term. Start with t1 then go to
t4 plugging in the previous term for (tn-1)

Find the sum of the first ten terms of the series:
2 - 6 + 18 - 54 +...
s10 = 2(1 - (-3)^10) / 1 - (-3)
s10 = -29, 524

*(2 being the first number in the problem, -3 being what you multiply each number by
to get the next term)

Find the sum of the first 25 terms of the arithmetic series:
11 + 14 + 17 + 20 +...
tn = 11 +(25 - 1)3
tn = 83
s25 = 25(11 +83) / 2
s25 = 1175

*(11 being the first number in the sequence, 3 being the number you add, Plug 25 into the
n-1 formula because your looking for the 25th term)

One of the MANY late reflections.

So, obviously I'm definitely behind on this blog thing...
I have NO idea what I've been doing! On a sunday, i know i have to do them then i wake up Monday morning and think "oops"...


So here's one of four:


I'll discuss some of the formulas and trig functions we've learned recently.
Cos(α +/- β)=cos α cos β -/+ sin α sin β
sin(α +/- β)=sin α cos β -/+ cos α sin β
sin x + sin y= 2 sin x + y/2 cos x-y/2
sin x - sin y= 2 cos x + y/2 sin x-y/2
cos x + cos y= 2 cos x + y/2 cos x-y/2
cos x - cos y= 2 sin x + y/2 sin x-y/2

tan (α + β)=tan α + tan β/1-tan α tan β
tan (α - β)=tan α - tan β/1+tan α tan β

sin2α=2sin α cos α
cos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1
tan 2α = 2tan α /1-tan^2 α
sin α/2= +/- √1-cos α/2
cos α/2= +/- √1+ cos α/2
tan α/2= +/- √1-cos α or 1 + cos α
=sin α/1+cos α
=1-cos α/sin α

something that i understood the most was section 2
here's some examples:

tan α = 2 and tan β=1
find tan (α - β)
= tan α + tan β/1-tan α tan β
=2+1/1-(2)(6)
=3/-1
=-3

Find the exact value of: tan 15+tan 30/1-tan 15 tan 30
tan α = 2 and tan β=1
find tan (α - β)
= tan (15 + 30)
=tan (45)
=1


Oh, and did I mention that I failed my trig exam MISERABLY? :/
& i actually studied!

5/9

Here is how you change the base of a log.

Change of Base:

Steps:

1. Take the log (base what you want of both sides)
2. Write as an exponential
3. Move exponent to the front
4. Solve for variable
5. Write as a fraction or whole number if possible. If not possible leave in log form

Ex: 1. 5^x = 10
2. log 5^x = log 10
3. xlog 5 = 1
4. Answer: x = 1/log 5

Last Reflection

logb^x=a
all you would do is switch the x and the a. (Kinda like an inverse)
b^a=x <---This is exponential form.

log3^2=7
3^7=2

This helps you solve problems sometimes.
But before you solve you have to change it to exponential form.

Example:
logx^4=2
x^2=4
x= +/- 2

For natural logs-
e^x=75
change e to ln(natural log), then switch the x and the 75
ln75=x
Plug into your calculator.
ln 75=4.317

reflection

This week was pretty fun, dressing up in different clothes everyday for pi week. It went by kind of slow..It started off with the trig exam on monday and tuesday, which i thought i did pretty good on, and come to find out i did to pretty good:)..Then for the rest of the week we just worked on our chapter tests for the final exam in a few weeks. The tests are chapters 1-6 and 13, if anyone didn't know that. On this reflection im going to review parabolas.....

REMEMBER:
1) x=-b/2a axis of symmetry

2) 2 ways to find vertex

(-b/2a, f(-b/2a)) or comeplete the square to get vertex form.

y=(x+z+^2 +b a&b are numbers
(-a,b)---vertex

3) focus:
1/up = coeff. of x^2
then add p

*if opens up add to y-value from vertex
*opens down, subtract
*opens right add to x-value
*opens left subtract

4) directrix is p units behind vertex
always x= or y=
so subtract p

**directrix is a line, focus is a point!
**if opens up subtract & is opens down add; from y-value of vertex
**opens right subtract x-value
**opens left add x-value

EXAMPLES:

1) 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)

D: y=1-1/4
y=3/4

Reflection

I'll explain finding the symmetry and domain and range since that will be on our final exam.

finding the symmetry:

EXAMPLE: y=x^3=4x

a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No

b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No

c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No

d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes

Domain and Range:
These are domain and range problems with fractions and polynomials.

EXAMPLE: y=x^3+4x^2+12
For any type of polynomial the domain would be (-oo,oo) and for the range, odd:
(-oo,oo); quadratics: [vertex,oo] or [-oo,vertex].
The answer to this problem would be Domain: (-oo,oo) Range (-oo,oo)

EXAMPLE: 5x+4/x^2-4
First, set the bottom of the fraction equal to zero.
x^2-4=0
+4+4
x^2=4
x=+ or - 2
Your answer then comes to,
Domain: (-oo,-2)u(-2,2)u(2,oo)

reflection 4/9

WOOOOOOOOOOOOOOO! 3 more weeks of mathh! which means 3 more weeks of blogg. which means a whole nother year of calculus:( For people who did bad doing trig, here are some following trig stufffffffffffffff...


SOHCAHTOA.

sin=opposite leg/hypotenuse
cos=adjacent leg/hypotenuse
tan=opposite leg/adjacent leg
*You use SOCAHTOA for right triangles

Here are some examples on using SOHCAHTOA....

solve for b & c.

1) You have an ABC triangle with angle A 28 degrees, a=40, and angle C is a right triangle.
(The first thing you would do is draw out this problem and label all your sides and angles if they are given. Then look at this problem and see which angles or sides go with what, sin, tan, or cos. In this problem it's tan because once you draw it out 40 is across from you angle and C is adjacent.)
It's tan because
tan28=40/b
btan28=40
b=40/tan28

which equals about 75.229




word problem
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



see everyone tomorrow!!!!

reflection may 9

This week was pretty fun, dressing up in different clothes everyday for pi week. It went by kind of slow..It started off with the trig exam on monday and tuesday, which i thought i did pretty good on, and come to find out i did to pretty good:)..Then for the rest of the week we just worked on our chapter tests for the final exam in a few weeks. The tests are chapters 1-6 and 13, if anyone didn't know that. On this reflection im going to review parabolas.....

REMEMBER:
1) x=-b/2a axis of symmetry

2) 2 ways to find vertex

(-b/2a, f(-b/2a)) or comeplete the square to get vertex form.

y=(x+z+^2 +b a&b are numbers
(-a,b)---vertex

3) focus:
1/up = coeff. of x^2
then add p

*if opens up add to y-value from vertex
*opens down, subtract
*opens right add to x-value
*opens left subtract

4) directrix is p units behind vertex
always x= or y=
so subtract p

**directrix is a line, focus is a point!
**if opens up subtract & is opens down add; from y-value of vertex
**opens right subtract x-value
**opens left add x-value

EXAMPLES:

1) 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)

D: y=1-1/4
y=3/4

Reflection 5/9

Okay, so I liked this week overall, besides the fact that we took the two trig exams, which were fairly easy and i got a B so i'm happy. Happy Mothers' Day to all you mothers out there and B-Rob of course. Now, I'm going to discuss some limit stuff we learned a couple of chapters back.

Limits are denoted as lim with and n (arrow) infinity underneath it.

Here are the rules:

1. lim(n infinity) - if the degree of the top = the degree of the bottom then the answer is the coefficients

2. If the degree of the top is greater than the degree of the bottom, then it is infinity

3. If the degree of the top is smaller than the degree of the bottom, then it is 0.

IF the rules don't apply, you will have to use your calculator to find what the sequence is approaching.

For a geometric sequence if the absolute value of r is less than 1, then it goes to 0.


Examples:

lim (n infinity) sin (1/n)
sin(1/100) = .010
sin(1/1000) = .0010
sin(1/10000) = .00010


lim (n infinity) n+5/n = 1, because the degree is the same, so the coefficients equal 1


I'm also having some trouble with trying to simplify equations that deal with the trig functions. I guess I just need to learn my formulas more and know how to apply them to different types of equations and problems. After all, I'm going to know how to do this 100x more next year for Calculus next year, I'm guessing. So can some people give me some examples on how to do the simplifying with these problems? Thanks.

Reflection

This week we took our 2 trig test and worked on review packets for other sections. I find chapter 11 is pretty simple. I understood 11-3 and 11-4. We learned stuff about cardioids, limacons, rose, petals, archimedes spirals, and logmarithmisc spirals. The quiz on that was easy because its simple to understand the formulas with the graphs.

EXAMPLES FROM 11-4 AND 11-3

1)re^i(theta)=rcis(theta)
3e^i(2pi)=3cis2pi
=3cos2pi+3sin2pi(i)
=3(1)+3(0)i
=3

2)De Moivre's Thereom
z^n=r^ncis(n)(theta)
z=2cis20degrees Find z^2
z^2=2^2cis2(20degrees)
z^2=4cis40degrees

3)De Moivre's Thereom
4cis15degrees Find z^4
z^4=4^4cis4(15degrees)
z^4=256cis60degrees

4)Find z^1/4, z=16cis180degrees
z^1/4=16^1/4cis(180*/4+0X360*/4)= 2cis45*
z^1/4=16^1/4cis(180*/4+1X360*/4)= 2cis135*
z^1/4=16^1/4cis(180*/4+2X360*/4)= 2cis225*
z^1/4=16^1/4cis(180*/4+3X360*/4)= 2cis315*

5)Find z^1/2, z=9cis60*
z^1/2=9^1/2cis(60*/2+0X360*)= 3cis30*
z^1/2=9^1/2cis(60*/2+1X360*)= 3cis210*

REFLECTION 5/9

Only a few more blogs left you guysss!! I'm soooo excited haha. Well anyway, this week we started the review for our exam, which is Chapters 1 through 6 and Chapter 13. I'm just going to go over a few things from Chapter 4 this week. So here's a few examples:

Ex. 1) -2x^2 + 3x - 6 Find the vertex.
*To find the vertex all you have to use is the axis of symmetry formula which is:
x = -b/2a and you plug in what you get for that for the y coordinate
In other words, this is what the vertex would be:
(-b/2a,f(-b/2a))
*So when you plug into -b/2a you get: x = -3/2(-2) So x = 3/4
Now you plug 3/4 into the equation to get the y coordinate
So you get: -2(3/4)^2 + 3(3/4) - 6
= -9/8 + 9/4 - 6
= -39/8
**So your vertex is (3/4,-39/8)

Ex. 2) y = x-2x Find the inverse and prove that it's an inverse.
*Okay first of all, you have to type this function into your calculator to see the graph. And if the graph passes the horizontal line test (the line only touches the graph at one point) then the function has an inverse.
*The graph is a line so it has an inverse
*Now to find the inverse you take the equation you're given, switch the x's and y's and solve for y.
*So you get: x = y-2y
x=-y
So the inverse is y = -x
Prove:
1. F(f^-1(x)) = (-x) - 2(-x)
= -x+2x
=x (*if you get "x" as the answer, then you know you're right)
2. F^-1(f(x)) = -(x-2x)
=-x+2x
=x

Ex. 3) f(x)=x+6 g(x)=3x+1 **Find g(f(x))
*All this function is telling you to do is to plug the f equation into the g equation every time you see an x
*So you get 3(x+6)+1
=3x+18+1
*So your answer is 3x+19

(:

Reflection #38

Since we started exam review, we're doing ch. 1-6 (and 13)...
I'm going to review the rational root theorem, one of the things I've remembered:

ex. x^4+2x^3-7x^2-8x+12
factors of p(constant)=+/-12, +/-1, +/-6, +/-2, +/-3, +/-4
factors of q(leading coeff.)=+/-1
Find p/q=+/-12, +/-1, +/-6, +/-2, +/-3, +/-4
Plug all possibilities into table on calculator and find all where y is 0.
You should get 1, 2, -2, and -3.
Do cythetic division with any two.
1 1 2 -7 -8 12
1 3 -4 -12
1 3 -4 -12 0
(x^3+3x^2-4x-12)(x-1)
2 1 3 -4 -12
2 10 12
1 5 6 0
(x^2+5x+6)(x-1)(x-2)
Solve by factoring x^2+5x+6=
(x+2)(x+3)
x=-2, -3, 1, 2
Put in point form:
(-2, 0)(-3, 0)(1, 0)(2, 0)

It's pretty simple stuff, but it's also easy to forget.

Reflection 5/2

Ah, the trig exam tomorrow and tuesday. Hopefully I'll do good and I've been studying a lot. Anyways, I'm going to explain some of the stuff from chapter 11 on polar form and rectangular forms.

11-2 Imaginary Numbers are no longer "imaginary"

Rectangular form is defined as
a + bi

Polar form is defined as
z = r cos theta + r sin theta i
abbreviated z = r cis theta

Example: Express 2 cis 50degrees in rectangular form

2 cos 50 + 2 sin 50 i


-1-2i in polar form
radius = +- sqrt of ((-1)^2 + (-2)^2)) = +- sqrt of (5)
theta = tan^-1(-2/-1)
theta = tan^-1(1)

if you do this, tangent is positive in the first and third quadrants, so it comes out to be 63.435 and 243.435

Since the 63 is positive for cosine, we can put it with the positive sqrt of 5.
And the 243 is negative for cosine, we can put it with the negative sqrt of 5.

z= sqrt of 5 cis 63.435
z= sqrt of 5 cos 63.435 + sqrt of 5 sin 63.435 i
z= negative sqrt of 5 cis 243.435
z- negative sqrt of 5 cos 243.435 + negative sqrt of 5 sin 243.435 i

It is now in polar form.

reflection

Trig test tomorrow..... aaaaaaaah. Hopefully I do good, anyways I'll review identites that we learned in chapter seven. The main thing to identities is memorizing all the formulas, if you don't then you wont be able to work the problem. To solve or simplify an identity you just replace what the have with something in the formulas.**Keep in mind you are trying to make the problem smaller not larger.

1. simplify:
(1-sinx)(1+sinx)
*Multiply the 2
1-sinx+sinx-sinx^2
=1-sin^2
*Use a formula
= cos^2x s^2+c^2=1
c^2=1-s^2
The answer would be 1.


2. prove:

cotA(1+tan^2A)/tanA=csc^2A

cotA(sec^2A)/tanA

=1/tan9sec^2A)/tanA

=(sec^2A/tanA)/(tanA/1) = sec^2A/tan^2A

=(1/cos^2A)/(sin^2A/cos^2A) = cos^2A/ cos^2Asin^2A = 1/sin^2A


= csc^2A

**For this problem you are just trying to get the same thing that you started off the problem with to show that it is correct. Solve the problem like you normally would and the answer should be what it equals up to.

Last Reflection

Now thinking way back to logs. Because this will probably help me since im not to great with them.

logb^x=a
all you would do is switch the x and the a. (Kinda like an inverse)
b^a=x <---This is exponential form.

log3^2=7
3^7=2

This helps you solve problems sometimes.
But before you solve you have to change it to exponential form.

Example:
logx^4=2
x^2=4
x= +/- 2

For natural logs-
e^x=75
change e to ln(natural log), then switch the x and the 75
ln75=x
Plug into your calculator.
ln 75=4.317

Reflection May 2

Today, i will explain Law of Sines.

ok here we go:

You are given that in triangle ABC:
B = 63 degrees, b = 5, and c = 4

Find angle C.

to solve for this you will use law of sines....
Set it up as:

sinC/4 = sin63degrees/5, because the formula is sinA/a = sinB/b.
Now cross multiply:
4sin63 = 5sinC
-divide by 5 to get sinC by itself:
sinC = 4sin63/5
-take inverse:
C = sin^-1 (4sin63/5) = 45.464 degrees
So angle C is 45.464 degrees, yay.



Now one with law of cosines:

You are given that in triangle ABC:
C = 120 degrees, b = 40.8, and a = 10.5

Find side c.

-follow the law of cosines formula, and plug in:
x^2 = 40.8^2 + 10.5^2 - 2(40.8)(10.5) cos120
-type into calculator with a square root symbol over all this, b/c its x^2
you get that c = 46.939 from your calculator.

reflection

Here are some identities:

Cos(α +/- β)=cos α cos β -/+ sin α sin β
sin(α +/- β)=sin α cos β -/+ cos α sin β
sin x + sin y= 2 sin x + y/2 cos x-y/2
sin x - sin y= 2 cos x + y/2 sin x-y/2
cos x + cos y= 2 cos x + y/2 cos x-y/2
cos x - cos y= 2 sin x + y/2 sin x-y/2

tan (α + β)=tan α + tan β/1-tan α tan β
tan (α - β)=tan α - tan β/1+tan α tan β

sin2α=2sin α cos α
cos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1
tan 2α = 2tan α /1-tan^2 α
sin α/2= +/- √1-cos α/2
cos α/2= +/- √1+ cos α/2
tan α/2= +/- √1-cos α or 1 + cos α
=sin α/1+cos α
=1-cos α/sin α

Does anyone have any helpful hints on this test so i dont fail?!?!?!

reflection 5/2

This has been a super long day, ive been studying math formulas all day and frustration and aggervation is kicking in, i feel very overwelmed, but i know i was taught all of this and hat i hopefully will do good on it the next two days.im not going to sit here and just type out formulas ill learn much better by keep on studying them right now. I will have better blogs in the future, i just think its more importat to concentrate on doing my math packets and studying my formulas.

Is there any helpful hints for remembering identites better!!!!!!!!!!!!!!!!

reflection march 2

So, this week was just like any other week. It went by pretty fast because softball left thursday to go to sulphur, and we didn't win:(..there's always next year. But anyway, we just looked over our tests and asked questions about the ones that we didn't know how to do and the ones that we are unsure about. This reflecton will be a little review over things that we learned in the past few weeks, before we started to review for the trig exam, which is tomorrow, and tuesday. That test is going to be the biggest thest i think i've ever taken. Im so scared:/...so anyway, heres a review on determinants.....


EXAMPLES:

a b c
d e f
g h i

*remember to delete the row, delete the column while alternating signs.
+,-+,-+,-

*pick row or column with the most 0's or 1's.

1. cross product - finds a pependicular vector
2. if uxv=0 then they are parallel.
3. /uxv/=/u/ /v/sin theta

u=(2,3,4) v=(1,0,5)

4. finds the area or volume, etc. of the object.

/uxv/

Reflection 5/2

okayyyyy, i'm super upset that i missed the field trip on friday, (haha NOT). But i am upset becasue i didn't get to study as much as i wanted to. I think i'm going to do okay on the exam though, because i understand all of the study guides from all of the previous chapters. This trig exam is going to be beast, but as long as i remember all of the identities, i should do okay, because i understand everything else that involves trig. So i've been stuck in my room all day reading over the study guides! Sounds boring, but as long as i get a good grade then it was worth it. So i'm just gonna give examples to review over before the exam to refresh myself and others.
_______________________________________________________________

Identities:

(Reciprocal relationships)

cscθ=1/sinθ
secθ=1/cosθ
cotθ=1/tanθ

(Relationships with negatives)

sin(-θ)=-sinθ
csc(-θ)=-cscθ
tan(-θ)=-tanθ
cos(-θ)=-cosθ
sec(-θ)=-secθ
cot(-θ)=-cotθ

(Pythagorean relationships)

sin^2θ+cos^2θ=1
1+tan^2θ=sec^2θ
1+cot^2θ=csc^2θ

(Cofunction relationships)

sinθ=cos(90-θ)
tanθ=cot(90-θ)
secθ=csc(90-θ)
cosθ=sin(90-θ)
cotθ=tan(90-θ)
cscθ=sec(90-θ)



EXAMPLE:

1) csc^2x(1-cos^2x)

1/sin^2x(sin^2x)=1

*both of the sin^2x cancel and you are left with 1.

2) cosθcot(90-θ)

cosθtanθ

cosθsinθ/cosθ=sinθ

*the cosθ cancel out and you are left with sinθ.

_______________________________________________________________

Formulas from chapter 10:

(Sum and difference formulas for cos & sin)

cos(a+-b)=cosacosb-+sinasinb
sin(a+-b)=sinacosb+-cosasinb

(Rewriting a sum or difference as a product)

sinx+siny=2sin x+y/2 cos x-y/2
sinx-siny=2cos x+y/2 sin x-y/2
cosx+cosy=2cos x+y/2 cos x-y/2
cosx-cosy=2sin x+y/2 sin x-y/2

(Sum formula for tangent)

tan(a+b)= tana+tanb/1-tanatanb

(Difference formula for tangent)

tan(a-b)= tana-tanb/1+tanatanb

(Double-angle and Half-angle)

sin2a=2sinacosa
cos2a=cos^2a-sin^2a...1-2sin^2a...2cos^2a-1
tan2a= 2tana/1-tan^2a
sin a/2= +-sqrt 1-cosa/2
cos a/2= +-sqrt 1+cosa/2
tan a/2= +-sqrt 1-cosa/1+cosa...sina/1+cosa...1-cosa/sina


EXAMPLE:

1) cos75cos15+sin75sin15

=cos(a+b)=cos60--> 1/2

2) sin75cos15+cos75sin15

=sin(a-b)=sin60--> sqrt3/2

_______________________________________________________________

So pretty much my whole blog i just typed is what i need to study the most. That is probably the case with everyone else too. All of the other trig stuff is easy. But if anyone else can give me examples of the hard identity ones, i would greatly appreciate it :) And if anyone knows that there is some more confusing problems on the exam that you get. I could study that too, haha. THANKSS :)

Reflection

Trig test tomorrow..... aaaaaaaah. Hopefully I do good, anyways I'll review identites that we learned in chapter seven. The main thing to identities is memorizing all the formulas, if you don't then you wont be able to work the problem. To solve or simplify an identity you just replace what the have with something in the formulas.**Keep in mind you are trying to make the problem smaller not larger.

1. simplify:
(1-sinx)(1+sinx)
*Multiply the 2
1-sinx+sinx-sinx^2
=1-sin^2
*Use a formula
= cos^2x s^2+c^2=1
c^2=1-s^2
The answer would be 1.


2. prove:

cotA(1+tan^2A)/tanA=csc^2A

cotA(sec^2A)/tanA

=1/tan9sec^2A)/tanA

=(sec^2A/tanA)/(tanA/1) = sec^2A/tan^2A

=(1/cos^2A)/(sin^2A/cos^2A) = cos^2A/ cos^2Asin^2A = 1/sin^2A


= csc^2A

**For this problem you are just trying to get the same thing that you started off the problem with to show that it is correct. Solve the problem like you normally would and the answer should be what it equals up to.

----------------------------
Something thats confusing me is the whole graphing thing with the amplitude and period and phase shift and all that. Help pleaseeeeee?

Reflection

I'm going to review how you solve the area of a non right triangle. This is what you use to get the area of a non-right triangle: A= 1/2 (leg) (leg) Sin (angle between).

Example: For Area say you have a triangle with:
A side length of 4
A side length of 5
And an angle of 30 degrees
Then plug it all in: A=1/2 (4) (5) Sin 30 degrees
A= 10 Sin 30 degrees which aproximately = 5.

Example: For another triangle:A side length of 3
A side length of 8
And an angle of 60 degrees
Then plug it all in: A= 1/2 (3) (8) Sin 60 degrees
A= 12 Sin 60 degrees which aproximately = 10.392

5/2

We are having a math party at chauvins right now and I decided to take a break. Now Im doing my blog and Im going to give the formulas for the chapter 9 limacon, cardiod, and rose curve.

Limacon:

r=a+b sin (theta)
r=a+b cos (theta)

Cardiod:

r=a+-b sin theta
r=a+-bcos theta

rose curve

r=a sin (ntheta)
r=a cos (ntheta)

Archimedes

r=a theta+b

Logarithmic

r=ab^theta

reflection 5/2

well, May is here
time for the last few weeks of school, then summer :D
just can't wait for it
trying to get away from a whole mess that came up last week, and im an idiot on completely different levels -.-

anyways.........diverging from that statement.........back to math.........

well, i can't really think of much, and the trig exam is coming up, i guess i'll just put up the trig chart


0 30 45 60 90
sin 0=0 π/6=1/2 π/4=√2/2 π/3=√3/2 π/2=1
cos "=1 "=√3/2 "=√2/2 "=1/2 "=0
sec "=1 "=2√3/3 "=√2 "=2 "=Ø
csc "=Ø "=2 "=√2 "=2√3/3 "=1
tan "=0 "=√3/3 "=1 "=√3 "=Ø
cot "=Ø "=√3 "=1 "=√3/3 "=0


thats about all ive got
deff gonna need this for the trig exam
hope we all do good on it
good luck everybody

REFLECTION 5/2

Okay this week I'll go over some things from Chapter 11 to help review for that trig exam. Here's a few examples:

Ex. 1) Convert (12,13) to polar
*First, when you convert to polar you have to get two answers. the answers should be in the form (r,theta) *where r is a number and theta is a degree
*First to get r, you have to use this formula: r = sqrt of x^2 + y^2
*So you get r = +/- sqrt of 12^2 + 13^2
= +/-sqrt of 313
*Now to get the angle you take the inverse of tangent
*So you get tan^-1(y/x) >> tan^-1(13/12)
*and your reference angle for that is approx. 47.291 degrees (tangent is positive, so you're looking for the angles in the 1st and 3rd quadrants) So you add 180 degrees to 47.291 and you get 227.291
*47.291 and 227.291 are your two angles. now you have to figure out which angle goes with which sqrt of 313 (positive or negative)
first plot the point (12,13)..it ends up in the 1st quadrant. sooo the positive sqrt of 313 goes with the angle in the 1st quadrant which is 47.291
*So your final answers should be: (sqrt of 313,47.291)(-sqrt of 313,227.291)

Ex. 2) Convert (3,0 degrees) to rectangular
*to convert to rectangular you have to use these two formulas:
x = rcos(theta) y = rsin(theta)
(where r is the number 3 and theta is 0 degrees..and you're finding a coordinate (x,y))
*So first let's find the x. So you get x = 3cos(0)
= 3(1)
So x = 3
*Now find the y. So you get y = 3sin(0)
= 3(0)
So y = 0
**So your answer is (3,0)

Ex. 3) If z1 = 5 + 4i and z2 = 2 - 5i ...Find (z1)(z2)
*So all (z1)(z2) means is that you have to multiply the two together
*so you get (5+4i)(2-5i)
All you have to do is FOIL it out
So you get 10-25i+8i-20i^2
= 10-17i+20
So your answer is 30-17i

Ex. 4) Convert r = sin(theta) to rectangular
*okay when you have an equation like this and you're asked to convert, the first thing you have to do is use identities to get rid of the number in front of theta..but since there is no number there, you skip that step.
*Next you have to replace "sin" with "y/r" in the equation
*So you get >> r = y/r
now you have to get rid of the fraction, so you multiply both sides by "r".
*So now you have > r^2 = y
*Now you have to replace "r" with "sqrt of x^2+y^2"
So you get (sqrt of x^2+y^2)^2 = y ..then solve for y
so the square root and ^2 cancel out ^^
**So your final answer would be y = x^2 + y^2

***CAN SOMEONE PLEEEAAAAASSSEEEE HELP ME WITH NUMBER 5 ON THE CHAPTER 9 WORD PROBLEM TEST ASAP? (: ..I completely forgot how to do it.
It says:
Find the area of a 60 sided figure inscribed in a circle with radius 2cm

Reflection

Well this week ill post some stuff about logs because its something simple, so i can not be so stressed about trig. I'll just give some examples of what i thought was the easiest thing we learned this year: logs.

EXAMPLES:
Condensing.

1.)logm + log7 + 4logn
= log7mn^4

2.)5loga + logd + log6
= log6da^5

3.)logn - 3logh -logy
= n/yh^3

4.)4logt - logc
= t^4/c

Expanding.

1.)log5gh^2
= log5 + 2logh +logg

2.)m^3b^7/f
= 3logm + 7logb - logf

Reflection #37

Again, I'm going to review some trig that's going to be on next weeks Trig Exam:

First, all of the formulas we learned:
** a = alpha B = beta

cos (a+B) = cos a cos B - sin a sin B
cos (a-B) = cos a cos B + sin a sin B
sin (a+B) = sin a cos B + cos a sin B
sin (a-B) = sin a cos B - cos a sin B
sin x + sin y = 2sin (x+y/2) cos (x-y/2)
sin x - sin y = 2cos (x+y/2) sin (x-y/2)
cos x + cos y = 2cos(x+y/2) cos (x-y/2)
cos x - cos y = -2sin (x+y/2) sin(x-y/2)
tan (a+B) = tan a + tan B/1-tan a tan B
tan (a-B) = tan a - tan B/1+tan a tan B
sin 2a = 2sin a cos a
cos 2a = cos^2 a - sin^2 a
cos 2a = 1-2sin^2 a
cos 2a = 2cos^2 a -1
tan 2a = 2tan a/1-tan^2 a
sin a/2 = +/- square root of 1-cos a/2
cos a/2 = +/- square root of 1+cos a/2
tan a/2 = +/- square root of 1-cos a/1+cos a
tan a/2 = sin a/1+cos a
tan a/2 = 1-cos a/sin a

Ex: Find the exact value of sin 15 degrees
a=45 degrees B=30 degrees
sin (a-B) = sin a cos B - cos a sin B
sin (45-30) = sin 45 cos 30 - cos 45 sin 30
sin 15 = (square root of 2 over 2)(square root of 3 over 2) - (square root of 2 over 2)(1/2)
sin 15 degrees = (square root of 6 over 4) - (square root of 2 over 4)
= square root of 6 - square root of 2 all over 4

If tan a = 2 and tan B = -1/3 then find tan (a+B)
tan (a+B) = tan a + tan B/1-tan a tan B
tan (a+B) = 2 + -1/3 / 1-(2 x -1/3)
tan (a+B) = 5/3 / 5/3
= 1

2cos^2 10 -1
Use...cos 2a = 2cos^2 a -1
cos 2(10)
=cos 20 degrees

Now, conversions:

To convert degrees to radians: multiply by pi/180
ex. 120 degrees
120 times pi/180
(keep answer in pi form)
120 divided by 180
=2/3
your answer =2/3 pi or 2 pi/3

To convert radians to degrees: multiply by 180/pi
ex. 3 pi/4 or 3/4 pi
3 pi/4 times 180/pi
pi cancels out
3/4 times 180
3 times 180
=540
divided by 4
your answer =135 degrees

To convert degrees to radians, it has to be in decimal form:
119.2
37.92
and so forth.
119.2/180 (plug in cal.) 149/225
then multiply pi =149 pi/225

To convert degrees decimals to minutes and seconds:
take the decimal of the degrees and multiply by 60 for minutes
68.33 .33 times 60 =19.8
take the decimal of the minutes and multiply by 60 for seconds
19.8 .8 times 60 =48
your answer =68 degrees 19 minutes 48 seconds

*if seconds have decimal, round to the 3rd decimal place
*if minutes don't have decimals, leave in degrees minutes

To convert degrees minutes seconds to decimals:
68 degrees 19 minutes 48 seconds
divide seconds (48) by 3600 =.013
divide minutes (19) by 60 =.317
add the two decimals =.33
your answer is 68.33

heres another late blog. its a review from chapter 13. Find the sum of the multiples of 3 between 1 and 1000*The first thing you want to do is find the smallest and largest multiple of three between those numbers.*The smallest multiple is 3 and the largest is 999*You can use the arithmetic formula for this problem like this:999=3+(n-1)(3)999=3nn=333So now that you have 'n' you can plug all your numbers into the sum formula to get the answer*So you get Sn=333(3+999)/2And that gives you 166833

late blog

this is one of my make up blogs. so heres somthing that im going to review from chapter 10. Simplify tan75-tan30/(1+tan75tan30)*When you first look at this problem you should notice that it is a formula*It's a formula for tan(a-b) **a=alpha;b=beta*So all you do is plug the two numbers you're given into tan(a-b) where the bigger number is alpha and the other number is beta (So a=75 and b=30)*So you get tan(75-30)= tan 45And they're asking you to simplify, so that means your answer should be a number*tan 45 is 1, so 1 is your answer

Topics

Write a blog on one of the following

1. Explain in detail the steps to graphing a trig function

2. Explain how to solve a non-right triangle with Law of Sines and Law of Cosines

3. Explain how to know what type of polar graph you have given an equation

Reflection #36

I'm going to review the trig we've been working on in class:

First, the trig chart:
0 = 0 degrees
pi/6 = 30 degrees
pi/4 = 45 degrees
pi/3 = 60 degrees
pi/2 = 90 degrees
sin 0 = 0
sin 30 = 1/2
sin 45 = square root of 2 over 2
sin 60 = square root of 3 over 2
sin 90 = 1
cos 0 = 1
cos 30 = square root of 3 over 2
cos 45 = square root of 2 over 2
cos 60 = 1/2
cos 90 = 0
tan 0 = 0
tan 30 = square root of 3 over 3
tan 45 = 1
tan 60 = square root of 3
tan 90 = undefined
cot 0 = undefined
cot 30 = square root of 3
cot 45 = 1
cot 60 = square root of 3 over 3
cot 90 = 0
sec 0 = 1
sec 30 = 2 square root of 3 over 3
sec 45 = square root of 2
sec 60 = 2
sec 90 = undefined
csc 0 = undefined
csc 30 = 2
csc 45 = square root of 2
csc 60 = 2 square root of 3 over 3
csc 90 = 1

Examples:
**a=alpha
4 sin pi/6 cos pi/6
=2 sin 2a
=2 sin 2(pi/6)
=2 sin(pi/3)
=2(square root of 3 over 3)
=square root of 3

cos^2 x/2 - sin^2 x/2
=cos 2a
=cos 2(x/2)
=cos x

Now, identities:
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x + cos^2 x = 1
1 + tan^2 x = sec^2 x
1 + cot^2 x = csc^2 x
sin x = cos (90 degrees - x)
tan x = cot (90 degrees - x)
sec x = csc (90 degrees - x)
cos x = sin (90 degrees - x)
cot x = tan (90 degrees - x)
csc x = sec (90 degrees - x)
tan x = sin x/cos x
cot x = cos x/sin x

Examples:
csc x - sin x
1/sin x - sin x/1
multiply sin x/1 by sin x/sin x
1/sin x - sin^2 x/sin x
1 - sin^2 x/sin x
cos^2 x/sin x
= cot x cos x

cot x + tan x
cos x/sin x + sin x/cos x
multiply cos x/sin x by cos x/cos x
multiply sin x/cos x by sin x/sin x
cos^2 x/cos x sin x + sin^2 x/cos x sin x
cos^2 x + sin^2 x/cos x sin x
1/cos x sin x
1/cos x times 1/sin x
= csc x sec x

sin x cot x
sin x/1 times cos x/sin x
sin x cancels out
= cos x

reflection

Cos(α +/- β)=cos α cos β -/+ sin α sin β
sin(α +/- β)=sin α cos β -/+ cos α sin β
sin x + sin y= 2 sin x + y/2 cos x-y/2
sin x - sin y= 2 cos x + y/2 sin x-y/2
cos x + cos y= 2 cos x + y/2 cos x-y/2
cos x - cos y= 2 sin x + y/2 sin x-y/2

tan (α + β)=tan α + tan β/1-tan α tan β
tan (α - β)=tan α - tan β/1+tan α tan β

sin2α=2sin α cos α
cos 2α=cos^2 α –sin^2 α = 1-2 sin^2 α= 2 cos^2 α -1
tan 2α = 2tan α /1-tan^2 α
sin α/2= +/- √1-cos α/2
cos α/2= +/- √1+ cos α/2
tan α/2= +/- √1-cos α or 1 + cos α
=sin α/1+cos α
=1-cos α/sin α

tan α = 2 and tan β=1
find tan (α - β)
= tan α + tan β/1-tan α tan β
=2+1/1-(2)(6)
=3/-1
=-3

Find the exact value of: tan 15+tan 30/1-tan 15 tan 30
tan α = 2 and tan β=1
find tan (α - β)
= tan (15 + 30)
=tan (45)
=1

Reflection

Solving anything bigger than a quadratic usuing quadratic form. To use quadratic form you have to have 3 terms only. The first term must equal the 2nd exponentx2, and the last term must be a constant. The first thing you do is make g=x^exponent/2 so that you would get g^2+g+#. The second thing is to do the quadratic formula, factor, or complete the square. The the last thing is to plug back in for g. (Whenever you do step three you are basically just plugging back into step one. g=x^2)

An example:
x^4-4x^2-12=0

1. g=x^4/2
g=x^2
g^2-4g-12

2. (g^2-6g)+(2g-12)
g(g-6)+2(g-6)
(g+2)(g-6)
g=-2 g=6

Triangle Formulas:

A= 1/2 (leg) (leg) Sin (angle between). This is used for non-right triangles.

For Area say you have a triangle with:

A side length of 4
A side length of 5
And an angle of 30 degrees

Then plug it all in: A=1/2 (4) (5) Sin 30 degrees

A= 10 Sin 30 degrees which aproximately = 5.


For another triangle:

A side length of 3
A side length of 8
And an angle of 60 degrees

Then plug it all in: A= 1/2 (3) (8) Sin 60 degrees

A= 12 Sin 60 degrees which aproximately = 10.392

This formula can only be done when you have two given lengths and at least one angle.

4/26

The Dot Product is something that we just learned in advanced math.
Here are the formulas and how you do it.

DOT PRODUCT:
(2,3) (1,4)
**Formula v1v2=x1(x2) + y1(y2)
Multiply:
2x1=2 + 3x4=12
2 + 12 = 14

MIDPOINT:formula m = (x1+x2/2, y1+y2/2, z1+z2/2)
Find the midpoint of (2,2,2) and (2,4,6)
2+2/2, 2+4/2, 2+6/2
=(2, 3, 4)

SPHERE:
A(8,-2,3) and B(4,0,7) are endpoints of a diameter. Find the equation of the sphere.

**MIDPOINT
C=(8+4/2, -2+0/2, 3+7/2)
C=(6,-1,5)

Reflection

For this week's blog I'll just review some things from Chapter 10. So here's a few examples:

Ex. 1.) If sinx=12/13 Find tan x/2
*Okay for this problem they're asking you to find tan x/2, and that has a formula:
sinx/(1+cos x) So you're going to use that formula to find the answer
*So first you have to draw a triangle from what you're given (sinx=12/13)
(So the height of the triangle is 12, the hypotenuse is 13, and the base is 5)
*Now you can plug into the formula:
tan x/2 = sin x/(1+cos x)
=12/13 / 1+(5/13)
= 12/13 / 18/13 then sandwich it
=156/234 = 2/3
*So tan x/2 is 2/3

Ex. 2.) Simplify tan75-tan30/(1+tan75tan30)
*When you first look at this problem you should notice that it is a formula
*It's a formula for tan(a-b) **a=alpha;b=beta
*So all you do is plug the two numbers you're given into tan(a-b) where the bigger number is alpha and the other number is beta (So a=75 and b=30)
*So you get tan(75-30)
= tan 45
And they're asking you to simplify, so that means your answer should be a number
*tan 45 is 1, so 1 is your answer

Ex. 3.) Simplify cos(60 + y) + cos(60 - y)
*For this problem you're going to have to use the cosine sum and difference formulas and you'll have to work the two parts of the problem separately
*So first you can solve cos(60 + y) Plugging that into the formula you get:
(cos60cosy - sin60siny)
^^And that simplifies to give you (1/2)cosy - sqrt of 3/2siny
*Now you can solve cos(60 - y)
So you get: (cos60cosy-sin60siny)
^And that simplifies to (1/2)cosy + sqrt of 3/2siny
Now you add everything together:
(1/2cosy - sqrt of 3/2siny + 1/2cosy + sqrt of 3/2siny
sqrt of 3/2's cancel out so you're left with 1/2cosy + 1/2cosy
And that equals 1cosy => cos y is your answer

Ex. 4.) If tan x = 1/2 and tan y = 2/3 Find tan(x+y)
*they're asking for tan(x+y) so all you have to do is plug the numbers into the tangent sum formula:
tna(a+b) = tanz+tanb/(1-tanatanb)
*so you get 1/2 + 2/3 / 1-(1/2)(2/3)
= 7/6 / 1-1/3
=7/6 / 2/3 sandwich it and you get 21/12
*So your simplified answer is 7/4

Reflection 4/25

so, i guess im awake enough to do this now XD
ill do it on the unit circle and values of the functions

Unit Circle:

90°, (0,1), π/2
360°, 0°, (1,0), 2π
180°, (-1,0), π
270°, (0,-1), 3π/2

sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y

now, knowing all of this can be verrrryyy useful

if you see something like sin(8π/5), you can use basic knowledge to kno 2 things: a) that the angle is in the 4th quadrant b) that the angle's going to have a negative y coordinate and r will be positive
now, you can use math to realize that this angle's measurement in degrees is 288°
then, you could get the reference angle, which is 72°, and then you just plug it into sin72, and then you get 0.254

well, hope this was informative, and i hope everybody had a good time at prom, i kno i did :)
o and one more thing, cant wait till tomorrow, my birthday :D
k, cya y'all

Some more review on my favorite thing in trig because its SUPER EASY.


SOHCAHTOA!!!!!!

sin=opposite leg/hypotenuse
cos=adjacent leg/hypotenuse
tan=opposite leg/adjacent leg
*You use SOCAHTOA for right triangles
Here are some examples on using SOHCAHTOA....
solve for b & c.
1) You have an ABC triangle with angle A 28 degrees, a=40, and angle C is a right triangle.(The first thing you would do is draw out this problem and label all your sides and angles if they are given. Then look at this problem and see which angles or sides go with what, sin, tan, or cos. In this problem it's tan because once you draw it out 40 is across from you angle and C is adjacent.)It's tan because
tan28=40/b
btan28=40
b=40/tan28
which equals about 75.229
sin28=40/c
c=40/sin28
which equals about 85.202

reflection april 25

This week went by extremely slow. All we did was go over our old tests and ask questions to prepare for the giant trig exam. This weekend went by pretty fast, won the second round of playoffs and going to statteee! And then rushing back home to get ready for prom which was fun. Now back to another week of school getting ready for the trig exam. Heres some stuff from chapters 7 and 8 that we did during the week.

--CHAPTER 7--

10) tan(cos-1(12/13))
for this problem you would draw a triangle and then use SOHCAHTOA
you have 13, 5, and 12 in your triangle. then you would get 5/12 as your answer because tan=opp/adj


17a) sin 626
_sin_
the first blank would be pos or neg and the second would be for the angle between 0 &90
626-360=266
266-180=86

--CHAPTER 8--

5) cosxcotx=2cosx

cosxcotx-2cosx=0
cosx(cotx-2)=0
cosx=0 cotx-2=0
x=cot-1 (2)
x=tan-1(1/2)

The key to solving this problem or problems like these is to move to one side, which is trig, and then factor.

2) cosx/secx+sinx/cscx

cosx/1/1/cosx + sinx/1/1/sinx = cos^2x+sin^2x = 1

Reflection 4/25

Okay, so this week was pretty good, but it went by too slow. Congrats to the softball team on going to state, and I'm sure everyone had a great time at prom and prom mania on Saturday night. The Trig exam is coming up, and hopefully all this trig will get back into my mind and I can understand it all. I'm going to review a couple of things in chapter 10 for this week's blog.

Sum and Difference Formula for Cosine

cos(alpha +- beta) = cos(alpha) cos(beta) -+ sin(alpha) sin(beta)

Find the exact value of cos 75
alpha = 45 beta = 30
cos(a + b) = cos(a)cos(b)-sin(a)sin(b)
cos 75 = (sqrt of 2/2)(sqrt of 3/2) - (sqrt of 2/2)(1/2)
cos 75 = (sqrt of 6 - sqrt of 2)/4


The only thing I'm gonna have trouble with is learning all of the formulas, and then maybe those problems where you have to edit the formula for example sin 4a = blank, or something like that...if there could be any example problems for this, that would be very much appreciated. Thanks. And also, i'm confused on when you have to find one angle, or two, or three...stuff like that.

Reflection 4/25

This had to be the funnest weekend everrr! The softball team had a game on saturday, which we won, and now we're going to STATE:). To add to that excitement, we also had prom on saturday night. It was nearly impossible for us to get ready in time after the softball game, but believe it or not, we made it. Prom mania was the GREATEST! There were sooo many prizes and it was sooo much fun! Anyway, back to math...right? So this trig exam? Its going to be hard or what? So i'm going to review for the trig exam, to help myself and others refresh on what is possibly on the exam. We have about one week left until we take the exam! Ouuu, yaaayy! Ha, but i'm pretty sure that everyone should do good on it. If you study and actually understand what we learned. But if you didn't pay attention in class, then you may have a problem. So anywayyy...
___________________________________________________________

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

_________________________________________________________

Unit Circle:

90 degrees, (0,1), pi/2
360 degrees, (1,0), 2pi
180 degrees, (-1,0), pi
270 degrees, (0,-1), 3pi/2

sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y

EXAMPLES:

1.) sin 2pi = y/r = 0/1 = 0
2.) cos 180 degrees = x/r = -1/1 = -1
3.) sec 3pi/2 = r/x = 1/0 = undefined

__________________________________________________________

So the trig exam is only on chapters 7-11? If so, you can just give me some identity examples! I STILL don't understand how to do those. I guess i'll just have to sit down one night and memorize ALLLL of them. That is the only way i'll learn how to do those type of problems. But if anyone has any hints or examples, pleaseee?! :) HOPE EVERYONE HAD FUN AT PROM!

REFLECTION 4/25

For this week's blog I'll just review some things from Chapter 10. So here's a few examples:

Ex. 1.) If sinx=12/13 Find tan x/2
*Okay for this problem they're asking you to find tan x/2, and that has a formula:
sinx/(1+cos x) So you're going to use that formula to find the answer
*So first you have to draw a triangle from what you're given (sinx=12/13)
(So the height of the triangle is 12, the hypotenuse is 13, and the base is 5)
*Now you can plug into the formula:
tan x/2 = sin x/(1+cos x)
=12/13 / 1+(5/13)
= 12/13 / 18/13 then sandwich it
=156/234 = 2/3
*So tan x/2 is 2/3

Ex. 2.) Simplify tan75-tan30/(1+tan75tan30)
*When you first look at this problem you should notice that it is a formula
*It's a formula for tan(a-b) **a=alpha;b=beta
*So all you do is plug the two numbers you're given into tan(a-b) where the bigger number is alpha and the other number is beta (So a=75 and b=30)
*So you get tan(75-30)
= tan 45
And they're asking you to simplify, so that means your answer should be a number
*tan 45 is 1, so 1 is your answer

Ex. 3.) Simplify cos(60 + y) + cos(60 - y)
*For this problem you're going to have to use the cosine sum and difference formulas and you'll have to work the two parts of the problem separately
*So first you can solve cos(60 + y) Plugging that into the formula you get:
(cos60cosy - sin60siny)
^^And that simplifies to give you (1/2)cosy - sqrt of 3/2siny
*Now you can solve cos(60 - y)
So you get: (cos60cosy-sin60siny)
^And that simplifies to (1/2)cosy + sqrt of 3/2siny
Now you add everything together:
(1/2cosy - sqrt of 3/2siny + 1/2cosy + sqrt of 3/2siny
sqrt of 3/2's cancel out so you're left with 1/2cosy + 1/2cosy
And that equals 1cosy => cos y is your answer

Ex. 4.) If tan x = 1/2 and tan y = 2/3 Find tan(x+y)
*they're asking for tan(x+y) so all you have to do is plug the numbers into the tangent sum formula:
tna(a+b) = tanz+tanb/(1-tanatanb)
*so you get 1/2 + 2/3 / 1-(1/2)(2/3)
= 7/6 / 1-1/3
=7/6 / 2/3 sandwich it and you get 21/12
*So your simplified answer is 7/4

Reflection

This week we are doing trig review and we will be doing it next week. I am trying to understand this stuff better and its getting there but i got to study harder. This geometry type stuff just isn't my thing. I did understand a good bit of the worksheet she gave us on friday though.

EXAMPLE:

C = 90 degrees
b = 7
c = 12

First i found angle A
cos A = 7/12
A = cos^-1(7/12)
= 54.315 degrees

Second i found angle B
sin B = 7/12
B = sin^-1(7/12)
= 35.685 degrees

Next i found length of a
tan 54.315 = a/7
(7) tan 54.315 = a/7 (7)
a = 7 tan 54.315
a = 9.747

Finally i found the area
A = 1/2 bh
= 1/2 (7) (9.747)
= 34.115

Overall, this is becoming easier but if anyone wants to explain this a little better to me, please feel free to do so!

Reference Angles are to be between 0° and 90°.
Find which quadrant the angle is in, determine whether the sign is positive or negative in the quadrant, and last subtract 180° until the angle is between 0° and 90°.


How To Move To Different Quadrants:
I to II= make negative and add 180°
I to III = add 180°
I to IV = make negative and add 360°
II to IV = add 180°

Find the reference angle using the chart or a calculator, find what quadrant you need to be in, and then move to that quadrant.

BLOG 4/21

Okay I'll explain how to find a reference angle since that seems like the easiest thing. First you have to remember that you are JUST finding an angle, with the trig function, and not an exact value or number. You have to see which quadrant the angle is in first, to determine whether the trig function (angle) will be negative or positive. And remember that a reference angle can ONLY be between zero and 90 degrees. And you have to subtract 180 (usually) to get that degree. Here are a few examples:

Ex. 1.) tan 625
*First thing you have to do is determine which quadrant 625 is in.
*Wellllll, it isn't in the 4 quadrants, so you have to subtract 360 degrees from it. And you get 265 degrees. Now you figure out which quadrant 265 is in to see if your trig function will be positive or negative.
*265 degrees is in the 3rd quadrant, where tangent is positive (y/x).
*And to get your reference angle you subtract 180 degrees from 265 and you get 85 degrees.
**So you're reference angle is tan 85 degrees

Ex. 2.) cos 125
*First, see what quadrant 125 is in. It's in the 2nd quadrant, and you're dealing with cosine so cosine is negative in the 2nd quadrant.
*125 isn't between zero and 90 degrees, so you have to subtract 180 degrees from it. (ignore that negative sign)
*And you get 55 degrees
*So your reference angle would be -cos 55 degrees

Ex. 3.) sin 845
*First since 845 is not in the quadrants, you have to subtract 360 degrees from it. You get 485 degrees, which is still not in the quadrants so you have to subtract 360 again. And you get 125 degrees
*So you find 125 in the quadrants and it's in the quadrant 2, so sine is positve there.
*Then you subtract 180 from 125 and get 55 (ignoring the negative)
**So your reference angle is sin 55 degrees

Blog Topics

The Second Blog topic for this week is to explain one of the following:

1. How to use the unit circle

2. How to solve for an angle

3.How to find a reference angle

late blog

Okay, I'm going to review synthetic division:
when you divide a polynomial by a numberEx. Divide p(x)=x^3+5x^2-2 by x+21st) solve x+2 for x...x=-2 (that's the number you're dividing by)2nd) begin by listing each number (if there is not a term, replace it with 0)Ex. -2) 1 5 0 -23rd) bring down the first numberex. -2) 1 5 0 -2...........14th) multiply 1 and -2 and place underneath second numberex. -2) 1 5 0 -2............-2...........15th) add the second number and the number underneath (5 and -2)ex. -2) 1 5 0 -2............-2...........1 36th) and repeat until doneex. -2) 1 5 0 -2............-2 -6 12...........1 3 -6 107th) take the finished results and form an equation (the last number is the remainder)ex. x^2+3x-6 R10**if the remainder is 0, then the number you divided by is a factorEx. p(x)=2x^4+5x^3-8x^2-17x-6 divided by x-22) 2 5 -8 -17 -6......4 18 20 6...2 9 10 3 02x^3+9x^2+10x+3

Reflection 4/18

This week we learned all of Chapter 12 and are now taking a Chapter 12 Take-Home test.

-Dot Product
(2,3) (1,4)
Formula v1v2=x1(x2) + y1(y2)
Multiply:
2x1=2 + 3x4=12
2 + 12 = 14

-Midpoint
formula m = (x1+x2/2, y1+y2/2, z1+z2/2)
Find the midpoint of (2,2,2) and (2,4,6)
2+2/2, 2+4/2, 2+6/2
=(2, 3, 4)

-Sphere
A(8,-2,3) and B(4,0,7) are endpoints of a diameter. Find the equation of the sphere.

-Midpoint
C=(8+4/2, -2+0/2, 3+7/2)
C=(6,-1,5)

-Diameter
D= squareroot (4-8^2 + o-2^2 + 7-3^2)=squareroot 36 D=6/2



I'm finally finished my test, and i definitely can't afford to do bad.
so HOPEFULLLLLLY, i do well. :)

Blog 4/18/10

Yea so we are now in the 4th quarter working on chapter 12. Chapter 12 is working alot with vectors. Im understanding the stuff im just having trouble remember ing the formulas are how things work, im understanding just not maintaining what im learning. I'm aslo super worried about the trig exam. I know the stuff cause i have seen it, im just worried that i wont remember what to do when it comes time for the test. i hope i prepare myself enough. Anyway this week we did stuff in chapter 12 dealing with vectors.

DOT PRODUCT:
(2,3) (1,4)
**Formula v1v2=x1(x2) + y1(y2)
Multiply:
2x1=2 + 3x4=12
2 + 12 = 14

MIDPOINT:formula m = (x1+x2/2, y1+y2/2, z1+z2/2)
Find the midpoint of (2,2,2) and (2,4,6)
2+2/2, 2+4/2, 2+6/2
=(2, 3, 4)

SPHERE:
A(8,-2,3) and B(4,0,7) are endpoints of a diameter. Find the equation of the sphere.

**MIDPOINT
C=(8+4/2, -2+0/2, 3+7/2)
C=(6,-1,5)

**DIAMETERD= squareroot (4-8^2 + o-2^2 + 7-3^2)=squareroot 36 D=6/2

I hope i did good on that take home test too haha

o to solve a determinate this is what you do

-Delete the row; delete the column while alternating signs. +,-,+,-,+,-...
-Pick row or column with the most zeros or ones.

Reflec

Vectors.

DOTS OF VECTORS

U(2,5) V(3,2) Find U dot V
(2)(3)+(5)(2)= 16

ADDING AND SUBTRACTING VECTORS
U+V= 2+3,5+2=(5,7)
U-V= 2-3,5-2=(-1,3)

MAGNITUDE OF A VECTOR
|U|= SQ. ROOT OF 2^2+5^2= SQ. ROOT OF 29

Reflection 4/18

This week in Advanced Math we learned all about vectors and how to find the vectors. We also learned how to find the determinant of 3x3,4x4, and 5x5 matrices. We also still had to finish our packet we got before the easter holidays. Then on thursday we were given a chapter 12 take home test.

On the take home test how do u work number 12?


How do you find the area of the paralellogram determined by the two vectors (2,5,9) and (2,7,1)


Also number 11?


How do you find a vector perpendicular to the two vectors (2,5,9) and (2,7,1)

I dont have a clue how to do these two problems.

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

Someone help me out with determinants??

Reflection 4/18

So, this week seemed to drag on a bit, but at least we only had like 2 periods on Friday. This week we continued to learn about vectors and other stuff relating to them. I will explain how to find the determinant of a 3x3 matrix and give an example.

To find a determinant of a 3x3 matrix (relating to vectors):

| a b c |
| d e f | =
| g h i |

-Delete the row; delete the column while alternating signs. +,-,+,-,+,-...
-Pick row or column with the most zeros or ones.

| a b c |
| d e f | = a| e f | - b| d f | + c| d e |
| g h i | | h i | | g i | | g h |

a(ei-hf) - b(di-gf) + c(dh-ge)


| 1 2 3 |
| 4 5 6 | = 0| 2 3 | - 0| 1 3 | + 1| 1 2 |
| 0 0 1 | | 5 6 | | 4 6 | | 4 5 |

take the zero matrices out

and the determinant would be just c(dh-ge) = 1(5-8) = 1(-3) = -3


Leaving space is also an option when solving:

| o l e |
5| f o g | - +
| o o h |

5(o| l e | - 0| | + h| |)
( | o g | | | | |)


I'm still kind of confused on how to find if two vectors are parallel.

Reflection

This week we learned some new stuff. After learning about vectors and what they were, we started to learn 4 methods of solving them. We also learned about the dot product, which is just plugging numbers into a formula and multiplying, then adding. On a problem it wont tell you to use to dot product you will just have to know when to use it. We also learned how to use midpoint, midpoint involves 3 points, it's exactly like the formula we learned at the beggining of the year except letter z is added. The answer for midpoint will be 3 points after plugging in the numbers and solving. We also learned about equations of spheres. When it ask for the diameter or center you are just plugging into formulas, diameter is distance formula with z added, and center is midpoint. Center will come out to points, and Diameter will come out to a number.

DOT PRODUCT:
(2,3) (1,4)
**Formula v1v2=x1(x2) + y1(y2)
Multiply:
2x1=2 + 3x4=12
2 + 12 = 14

MIDPOINT:
**formula m = (x1+x2/2, y1+y2/2, z1+z2/2)
Find the midpoint of (2,2,2) and (2,4,6)
2+2/2, 2+4/2, 2+6/2
=(2, 3, 4)

SPHERE:
A(8,-2,3) and B(4,0,7) are endpoints of a diameter. Find the equation of the sphere.

**MIDPOINT
C=(8+4/2, -2+0/2, 3+7/2)
C=(6,-1,5)

**DIAMETER
D= squareroot (4-8^2 + o-2^2 + 7-3^2)
=squareroot 36
D=6/2