6-4 Hyperbolas

This week we learned section 6-4. Section 6-4 was dealing with hyperbolas

Hyperbolas are in the form x^2/a^2 –y^2/b^2 or –x^2/a^2 + y^2/b^2.

Part of the Equation:
1. Major axis – variable with larger denominator (non-negative)
2. Minor axis –variable with smaller denominator
3. Length of major –2 square root of non-negative denominator
4. Length of minor –2 square root of smaller denominator
5. Vertex – square root of non-negative denominator
If x is major: (_,0) and (-_, 0)
If y is major: (0,_) and (0, -_)
6. Other intercepts – square root of smaller denominator … ( , ) opposite of vertex
7. Focus – focus squared = larger denominator + smaller denominator
If x is major: (focus, 0)
If y is major: (0, focus)
8.Asymptotes –
If x is major: y = +/- b/ax
If y is major: y = +/- a/bx


Examples:

X^2?25-Y^2/100=1

1. x
2. y
3. 10
4. 20
5. (5,0)(-5,0)
6. (0,10)(0,-10)
7. (5square root of 5,0)( -5square root of 5,0)
8. y= + and – 2x


Basically if you know your rules and stuff it easy and it the same thing as 6-3 rules but couple new steps to 6-4 that all.

6-4

Sooo I hope everyone had an awesome time at prom this weekend. I can’t remember if we did hyperbolas this week cause we only had one day of learning this week. The rest of the week we did aleks. So I’m just gonna do section 6-4 hyperbolas.

In the form x2/a – y2/ = 1 or –x2/a2 + y2/b2/

Parts of the equation
-Major axis : variable with larger denominator
-Minor axis: variable with smaller denominator
-Length of major: 2 square root non negative denominator
-Length of minor: 2 square root smaller denominator
-Vertex: square root non negative denominator if x is major (_,0 )and (-_,0); if y is major (0,_) and (0,-_)
-Other intercept: smaller denominator ( , ) Opposite of vertex
- Focus: focus2 = larger denominator and smaller denominator if x is major (focus,0) if y is major (0,focus)
-Asymtoles: y = +/-b/ax if x is major, y = +/-a/bx if y is major

solving the system

Okay, so this might be a little short because there was only one thing we learned this past week because we did aleks for the rest of the days. The 28 aleks that we are supposed to do are going to be very difficult to finish since we didn’t get to work in the library on Thursday. Once im done with this, Ill probably be working on that until I fall asleep. Also with prom this weekend, this is the first chance I am having to start on my schoolwork. I hope everyone had a great time (: On Wednesday, we learned something from algebra II called solving the system.

Honestly, I tried a bunch of different times to work a problem to put on here but all of the different numbers I tried just would not work out. They all came out with big fractions that I just don’t know how to complete. Im sorry I couldn’t figure it out but it is a pretty simple process whenever you already have a problem given to you. (This blog is over 150 words though)

6-4

Prom is this weekend, tonight actually and I’m stressin’ big time! So this might be a little short..
Section 6-4 dealt with hyperbolas. Hyperbolas are in the form x^2/a^2 –y^2/b^2 or –x^2/a^2 + y^2/b^2.
Parts of the equation are similar to the ellipse.
1. Major axis – variable with larger denominator (non-negative)
2. Minor axis – variable with smaller denominator
3. Length of major – 2*square root of non-neg denominator
4. Length of minor – 2*square root of smaller denominator
5. Vertex – square root of non-neg denominator
If x is major: (_,0) and (-_, 0)
If y is major: (0,_) and (0, -_)
6. Other intercepts – square root of smaller denominator … ( , ) opposite of vertex
7. Focus – focus squared = larger denominator + smaller denominator
If x is major: (focus, 0)
If y is major: (0, focus)
**SOMETHING DIFFERENT: Asymptotes –
If x is major: y = +/- b/ax
If y is major: y = +/- a/bx

EXAMPLE 1: x^2/9 – y^2/16 Find all parts.
1. x (since 16 is negative)
2. y
3. 2*square root of 9 = 6
4. 2*square root of 16 = 8
5. square root of 9 = (3, 0) and (-3,0)
6. square root of 16 = (0,4) and (0,-4)
7. 9 + 16 = 25
focus = 5
8. y = +/- 4/3x

Hope everyone has a fun, memorable weekend!

a matrices review..

Prom is next saturday:D but im sure you guys already know :) im kinda excited about it too. But anyways,I cant seem to find my chapter six notes so ill blog about something we did a while back. Matrices, yayyy. Just like every other area of advanced math, matrices are tedious.

Here are the matrices rule:

When adding matrices, they must be of the same dimensions. The same goes for subtracting

Ex. A 4x3 matrices can only be added to a 4x3 matrices

When multiplying matrices, you can only multiply matrices that have the same columns and rows.

Ex. 3x3* 1x2 the column of the first are equal to the rows of the second.

You cannot technically divide a matrices, to "divide" a matrices, you simply multiply by the inverse..

finding the inverse.

0 1

1 0 = 0-1=-1 = 1/-1* 0 1

1 0 = 0 -1

-1 0

Zomg

Zomg. Look who finally remembered that blogs exist!

So this week in Advanced Math we started chapter six, but we skipped 6-1 because B-Rob said it was dumb. So, one thing we learned was section 2 which is all about connics and the equation of a circle.

First, you must remember how to complete a square. To complete a square, you must put it in standard form. Then, get rid of the coefficient of x^2 or y^2. Next, divide the middle term by 2 and square it. Add the number to both sides. Then factor in (x )^2 form.

The mid point formula is also needed for this. As a remindered, it's (Xv1+ Xv2)/2, (Yv1+Yv2)/2.

The distance formula is used as well when given a center point and an outside point.

The equation of a circle is (x-h)^2 +(y-k)^2 = r^2 where (h,k) is the center.

Example of how to make an equation--
C=(4,3) r=2
(x-4)^2 + (y-3)^2 = 4

Example of finding coordinates of the point where the lines intersect--
y= 2x-2 and circle x^2 +y^2=25
By calculator:
plug in y=2x-2 and y= + or - square root (25-x)^2
Then hit graph
Then hit second trace
The click a point on the top circle
Then a point on the line
Then guess where about they intersect.
Answer-
(3,4)

Chapter 6

Guess what we did this week? Yup, you guessed it, Math. We did Math this week, but not just any kind of Math, Advanced Math. and not just any old Advanced Math but Elipsises. Which, come to think of it, is very hard to pronounce in the plural sense of the word. Anyway, enough stalling just so I can reach the 150 word thing, no I would never do such a terrible thing.

Ellipses have many parts.

Major axis- Found by whcih axis has the larger denominator
Minor axis- Found by which axis has amaller denominator
Length of the major axis- Found with the equation. 2xSquare root(larger denominator)
Length of the minor axis-.............................................................(Smaller denominator)
Intercepts, major and minor- Is the square root(Larger denominator) or (smaller denominator). Takes place of corresponding variable, depending on which is major, the other variable is 0
Focus- I think its Larger Denominator=Smaller Denominator+Focus squared

This week in advanced math we started on chapter 6. In section three we learned about ellipses. Ellipses appear in the form of x2/a + y2/b = 1. Parts of the equation of ellipses are:

Major axis – the variable with the larger denominator
Minor axis – the variable with the smaller denominator
Length of the major axis – 2 square root of the larger denominator
Length of the minor axis – 2 square root of the smaller denominator
Vertex – square root of the larger denominator if x is major (_,0) and (-_,0) and if y is major (0,_) (0,_)
Other intercept – square root of smaller denominator ( ,) Opposite of vertex
Focus – Smaller denominator = larger denominator, focus2 is x is major (0, focus) (focus,0)

Example 1:
x2/9 + y2/36 = 1

1) Major axis - y
2) Minor axis – x
3) Length of Major – 2 square root 36 = 12
4) Length of Minor – 2 square root 9 = 6
5) Vertex – (0, 6) (0, -6)
6) Other intercept – (3,0) (-3,0)
7) Focus – 9 = 36 –f2 =-27 = square root 27 (0, square root 27) (0,-square root 27)

This past week we learned some of Chapter 6.

6-2
CONICS

*equation of a circle
(x-h)^2 + (y-k)^2 = r^2
**where (h,k) is the center.

D=square root of (y2-y1)^2 + (x2-x1)^2 = r

-get rid of coefficient of x^2 or y^2
-divide middle term by 2 and square it.
-add to both sides
-factor to (x)^2

EXAMPLE:
(x-7)^2 + (y+2)^2=25
CENTER: (7,-2)
RADIUS: square root of 25 = 5

6-3
ELLIPSES

* x^2/a + y^2/b = 1

PARTS OF THE EQUATION:
major axis-variable with larger denominator

minor axis-variable with smaller denominator

length of major-2 square root of larger denominator

length of minor-2 square root of smaller denominator

vertex-if x is major ( ,0) & ( ,0)
if y if major (0, ) & (0, )

other- opp. of vertex ( , )

focus- smaller denom. = larger denom. - focus^2
if x is major (0, focus)
if y is major (focus, 0)

6-3

This week we learned section 6-3. In 6-3 we learned how to solve problems in the form of x^2/a+y^2=1

Part of the Equations:

Major axis

Minor axis

length of major

length of minor

Vertex

Other Int

Focus


Examples: Find all the parts of X^2/4+y^2/25=1

1. Major axis= y
2. Minor axis=z
3 Length of major=10
4. length of minor=4
5. vertex= (0,5)(0,-5)
6. Other Int=(2,0)(-2,0)
7. focus=( 0,square root of 21)(0,-square root of 21)

Basically if you know all your steps and procedure you will not have any problem at this section and this will be a breeze for you to.

6-3

This week in advanced math we learned how to find parts of the equation. The equation has many parts and we learned how to identify them all. We also learned recently how to find the equation of the circle. This is done by finding the distance formula and plugging numbers in the right place into the equation. But in this we'll focus on 6-3 and finding parts of the equation.

In the form x^2/a + y^2/b = 1

Parts of the equation

x^2/9 + y^2/16 = 1

major axis- y

minor axis- x

length of major-2 the square root of 16 = 2(4)= 8

length of minor- 2 the square root of 9 = 2(3) = 6

vertex - (0,4) and (0,-4)

other int - (3,0) and (-3,0)

focus - 9-16 = f^2 -7=f^2 f= square root of -7

chapter 6

This week we did some pretty easy things in chapter 6 for the most part. The easiest thing for me personally is finding the 7 parts of an ellipse.
1. Tell the major axis
2. Tell the minor axis.
3. Length of major
4. Length of minor
5. Vertex
6. Other intercepts
7. Focus
All of these things are pretty easy to find.

Ex: x^2/25 + y^2/9 = 1
^^ all of them must equal 1.

1. Major axis- x because the number on the bottom is bigger
2. Minor axis-y
3. Length of major – 10 (2 times the square root)
4. Length of minor- 6
5. Vertex – (5,0) (-5,0) (square root of the major axis. The number goes in the x spot since it is larger.
6. Other intercepts- (0,3)(0,-3)
7. focus- 4 ( smaller denominator = larger denominator – f^2)

6-3

This week in advanced math we are learning how chapter 6 which is ellipses, conics, and hyperbolas. I don't find this very hard. All you need to do if follow the steps which you're teacher has given you. Today I will show you how to find an ellipse equation and all that good stuff.

**This is the form of an ellipse equation: x^2/a + y^2/b


Step or parts of ellipse equation:

1. Major axis is the variable with larger denominator.
2. Minor axis is the variable with smaller denominator.
3. Length of major is 2 times the square root of larger denominator
4. Length of minor is 2 times the square root of smaller denominator.
5. Vertex is square root of larger denominator.

If x is major you should use this (_,0) and (-_, 0).
If y is major you should use this (0,_) and (0, -_).

6. Other intercepts is square root of smaller denominator ( , ) opposite of vertex.
7. Focus is the smaller denominator = the larger denominator – the focus squared.

If x is major you should use this (0, focus).
If y is major you should use this (focus, 0).

EXAMPLE 1: Find all the parts of x^2/4 + y^2/16.
1. y
2. x
3. 8
4. 4
5. (0,4) (0,-4)
6. (2,0) (-2,0) 6. 7 = 16 – focus squared
focus = (0,3) (0,-3)

This is how you find the equation thingy or whatever to an ellipse.

6-3

6-3 Ellipses
This past week we learned about conics, ellipses, and hyperbolas.
Ellipses are in the form: x^2/a + y^2/b

Parts of the equation:
1. Major axis – variable with larger denominator
2. Minor axis – variable with smaller denominator
3. Length of major – 2*square root of larger denominator
4. Length of minor – 2*square root of smaller denominator
5. Vertex – square root of larger denominator
If x is major: (_,0) and (-_, 0)
If y is major: (0,_) and (0, -_)
6. Other intercepts – square root of smaller denominator … ( , ) opposite of vertex
7. Focus – smaller denominator = larger denominator – focus squared
If x is major: (0, focus)
If y is major: (focus, 0)

EXAMPLE 1: Find all the parts of x^2/9 + y^2/25.
1. y
2. x
3. 2 * the square root of 25 = 10
4. 2 * the square root of 9 = 6
5. Since y is the major, (0,5) and (0,-5)
6. (3,0) and (-3,0) 7. 9 = 25 – focus squared
focus = +/- 4 therefore, (0,4) and (0,-4) is your focus.

5-5

Mardi Gras Break Blog #2-
Chapter 5-Section 5

Chapter 5, section 5 includes three formulas you must know. They are listed below:

The first formula involves:
A(t) = A (1 + r)t
A- is the orignal amount
r- is the rate
t- is amount of time

The second formula involves:
A(t) =A bt/k
A- is the original amount
b- whether you half, triple, or quaterly
t- is the time
k-the time it takes to get to b
** used with doubling

The thrid formula involves:
P(t) = P er/t
** used with continuosly

EXAMPLE:
You must choose which formula you use when given a problem. Example:You invest 3,000 at 16% intrest compounded continuously. How much money will you have after 2 years?

Plug into your calculator P = 3000e^(.16)(2) = $7.351.02

Exponents

This was the best break ever. I hope that everyone had a good Mardi Gra break. We took an exam on chapter 5. The easiest section in chapter 5 was when it was dealing with rational exponents.

Ration exponents:


Formulas:
A(t)=A0b^(t/k)
-A0 is starting point
-b is halfing, doubling, ect
-t is time and k is time it takes to get to b

Examples:
3rd root of square root of x= x^1/3

(16/100)^1/2= square root of 16/square root of 100 which can be simplified to 2/5

27^2x=3^2
(3^3)^2x=3^2
3^6x=3^2
6x=2
x=1/3

5-7

Mardi Gras Break Blog #1-
This holiday was very much needed! I am going to hate having to go back to school, but hey only two more weeks until Easter! I hope everyone had a good Mardi Gras! Before we left for break we closed out Chapter 5 and took our exam. This part of trig is getting kind of tricky, but if you follow the basic steps it can be easy.

Chapter 5- Section 7
There are two methods we learned in chapter 5, section 7; we were taught how to solve for a variable as an exoponent and also learned to change the base of a log.

** Take the log of both sides when solving for a variable as an exponent. After that is done take the exponent and bring it to the front of the equation.

EXAMPLE 1:
Solve 10^x=8
1. Take the log of both the left and right side.
log 10^x = log 8
2. Now, bring the exponent to the front .x log 10 = log 8
3. Solve for the variable, in this case x.X = log 8/log 10

EXAMPLE 2:
Solve 3^x=81
1. Take the log of both sides.log 3^x = log 81
2. Bring the exponent to the front.x log 3 = log 81
3.Solve for x.x = log 81/log 3D).
Your final answer is 4.

**Change of base formula:
log b^a:log x^a/ log x^b

EXAMPLE 1:
Change log 3^9 to base 4.
1.Using the formula, you get log 4^7/log 4^9

Mardi Gras

B-Rob i wish you would have given us prompts. My brain is fried from playing too much baseball to remember what we did in class the other week. I know we did stuff with logs, which is easy sometimes but hard other times. It tends to be easier when it is just a simple log compared to when it is a big equation.

like: log(little 8)64=2, thats easy

Something like this: log(little8)x^(4700+10t)-log(little x)10+58, tends to get kind of difficult.

chapter 5

Mardi Gras #2
Okay, so for my second blog of the break, I am pretty much going to do the same thing that I did in my previous blog. Like I said, chapter 5 did give me some trouble and I am definitely not looking forward to going back to school tomorrow. I enjoyed my break a lot and I’m sure the rest of you have too. Unfortunately it had to end. Alright back to math.

*I won’t be able to get the final answer because I’m dumb and left my calculator somewhere*

There are 3 formulas that we were given that we needed to know the difference between them and when to use each.

1. A(t) = a0 b^t/k * this formula is used when doubling, halfing, tripling, etc..
2. A(t) = a0 ( 1 + r)^t *used with regular problems with no “special” words
3. P(t) = p0 e^rt * use this one when you see words such as continuously or compounding

Since I don’t have my calculator and they are all worked the same way, I’ll just do one example.

Ex: Suppose a radioactive isotope decays so that the radioactivity present decreases by 23% per day. If 56kg are present now, find the amount that will be present 6 days from noow.

Since there are no key words, you will use formula number 2.
A(t) = 56 ( 1 + .23 ) ^6
You just plug that straight into the calculator to get your answer.

chapter 5

Mardi Gras #1
Before the mardi gras break we learned chapter 5 and took our 3rd nine weeks exam on it. That was probably the hardest exam I took out of all my classes. However, the problems really are very simple to do, they can just get a little confusing at times. Some of them hardly require much work either. I’ll just show a few random problems of things we learned.

The first example is going to be about exponential equations:
The formula for this is: f( x ) = a x b^x

Find the equation of the exponential if f(0) = 5 and f(2) = 20
F(x) = a x b^x
F(x) = 5 x b^x
20 = 5 x b^ 2
B^2 = 4
B = +/- 2

Other things that are very simple is changing a base given logb^a, you will end up with an answer in the term log x^a/ log x^b

Example: change log3 7 to base 5
Log5 7/log5 3

Mardi Gras Break Number 2

Soooo I’m back again writing my second blog of the break. Once again I will stay I am very bummed that break it over already. I have been super busy all week and it just flew by. I really wanted to aleks extra credit but I did not have any time this week so that sucks. Anyway I’m going to do so more examples from chapter 5 which has to do with like logs and formulas and other stuff. It’s easy enough I guess but I guess we will have to wait for exam grades to be posted to see how well I know my stuff.

Chapter 5 involves three main formulas:

Formula 1 : A(t) = A (1 + r)t
Formula 2: A(t) =A bt/k
Formula 3: P(t) = P er/t

You must choose which formula you use when given a problem.

Example:

You invest 2,000 at 6% intrest compounded continuously. How much money will you have after 3 years?

Plug into your calculator P = 2000e^(.06)(3) = $6371.02

Mardi Gras Break Blog Number 1

Hey everyone I hope everyone had a great break, I know I did but I am super bummed it’s over already. It seems like it flew by! Anyway the last week we were in school we learned chapter 5 and our exam we took on Friday was on chapter 5. I hope I did okay enough on it to pass (: Now let’s see if I remember how to do any of this stuff cause it’s been awhile…Rational Exponents look easy enough so let’s get started, here ya go!

Exponents that involve roots:

Example 1:

3 square root of 4 = 41/3

Example 2:

(9/25)1/2 = 91/2/251/2 = square root 9/ square root 25 = 3/5

Solve :

Example 3:

49 4x = 78
(72)4x = 78 Bases must be the same
78x = 78
8x = 8 Set exponents equal
X = 1

Okay I think this blog screwed up the way everything looks but your smart people and can figure it out for yourselves.

mardi gras blog #2

guess what everyone?! rascal flatts concert is sundayyy! which is why im posting early and not waiting last minute :)

i hope everyone had a great mardi gras holiday like i did although i was busy everyday!

FORMULAS:
logb MN=logbM + logbN
logb M/N=logbM - logbN (anything in the bottom is NEGATIVE)
logM^k=KlogM

to set 2 log equal set the insides equal.


**right side is expanded
**left side is condensed

EXPAND:

logbMN^2
logbM+logbN^2
=logbM+2logbN

CONDENSE:

log 45-2log3
log45-log3^2
log45-log9
log45/9
=5


byeeeeee peeps :)

Mardi gras blog #1.

so last week we completed chapter 5 and took our super hard exam. anddddd we did aleks.

chapter 5 was pretty decent, logs are not THATTTTTTT bad but i guess i didnt know them as good as i thought.

writing in exponentials.
EXAMPLE.
log4 64=3
4^3=64.
^^that's your answer. soooooo easy.

there's some with word problems but im not very good at making up word problems much less making them with numbers that will actually give me a normal answer :(

FORMULAS:
a(t)=a0(1+r)^t
a0----your starting point.
r----your decimal number
t----time ( must match rate )
(used with percents)

a(t)=a0 b ^t/k
(used when doubling, halfing, tripling)

p(t)=P0e^rt
p0---starting point
r---rate as decimal
t---time
Keywords (continuously, compounding)

5-6

Mardi Gras Break Blog #2:
5-6
Properties of Logs:
1. Log b MN = log b M + log b N
2. Log b M/N = log b M – log b N
3. Log b M^k = k log b M
(RIGHT SIDE IS CONDENSED, LEFT SIDE IS EXPANDED).

EXAMPLE 1: Condense log M – 3 log N
A). First, check to see which property it’s following. With the – sign, we automatically know the
answer will be in fraction form, property #2.
B). Also, the second half (3 log N) is property #3. Changing that, we get log N^3
C). Now we can continue..
log M/log N^3

EXAMPLE 2: Condense log 8 + log 5 – log 4
A). Simply the 3 parts to get 2 so we can condense. Since log 8 and log 5 are the same, multiply 8 and 5.
B). Now we have log 40 – log 4.
C). Having the – sign, we know we’re going to follow property #2.
log 40/log 4
D). Your answer is 10.

EXAMPLE 3: Expand log b MN^2
A). Following property #1, we get log b M + log b N^2
B). Now bring all exponents to the front to get your final answer.
log b M + 2 log b N

5-7

Mardi Gras Break Blog #1:
5-7
When solving for a variable as an exponent, you take the log of both sides. Then you bring the exponent to the front and solve.

EXAMPLE 1: Solve 10^x=4
A). Take the log of both the left and right side.
log 10^x = log 4
B). Now, bring the exponent to the front.
x log 10 = log 4
C). Solve for the variable, in this case x.
X = log 4/log 10

EXAMPLE 2: Solve 3^x=81
A). Take the log of both sides.
log 3^x = log 81
B). Bring the exponent to the front.
x log 3 = log 81
C). Solve for x.
x = log 81/log 3
D). Your final answer is 4.

To change a base given log b^a:
log x^a/ log x^b

EXAMPLE 1: Change log 3^7 to base 4.
A). Using the formula, you get
log 4^7/log 4^3

5-7

Last week we learned section 5-7. In 5-7 it was dealing with log and solves it by a variable as an exponent.

To solve for a variable as a exponent you take the log of both sides, bring the exponents to the front and solve.
- to change a base given log( base is b)^a

Log( base x)^a/log( base x)^b is where x is the base you want

Examples:


Change log (base 5)^8 to base 2

Log(base 2)8/ log (base 2) 5


Solve 3^x=81
Log 3^x=log 81

X log 3= log 81
X=log 81/log 3 = 4


If you know all your steps and the procedure of doing this you will not get stuck on this and it will be really easy for you.

Rational Exponents

This week we learned a section in chapter 5 which deals with rational exponents. This section is a easy and not that hard at all.

Formulas:
A(t)=A0b^(t/k)
-A0 is starting point
-b is halfing, doubling, ect
-t is time and k is time it takes to get to b

Examples:
3rd root of square root of x= x^1/3

(16/100)^1/2= square root of 16/square root of 100 which can be simplified to 2/5

27^2x=3^2
(3^3)^2x=3^2
3^6x=3^2
6x=2
x=1/3
Basically if you know your rules and the formula this section will be easy..

chapter 5

This week we learned things from chapter 5 and completed a lot of aleks modules. 5 was about exponents and a couple formulas. Unfortunately, exponents are not the easiest thing for me which is why I think I did so much worse in algebra II. Since I’m not very good at those, I’ll do a formula example.

The first formula we did was the easiest for me. It is:
A(t) = Ao (1 + r)^t

This formula is used with word problems when given a percent, time and rate.
Ao= your starting number
R= the rate
T= time

After you read the problem, you plug in the numbers to the formula, type it in your calculator, and it gives you the answer. (percents must be changed to decimals)

Ex: Suppose you have $1,000 dollars invested at a 7% interest rate. How much money will you have after 2 years?

A(t) = $1,000 (1 + .07)^2
^^ plug that into your calculator and you get $1144.90 as your answer.

Chapter 5

This week in advanced math we had to complete seven aleks modules and this weekend we have to do fourteen. I finished both so I’m super glad. Mrs. Robinson wasn’t here for a few days but we did manage to start chapter five. We were also supposed to get packets this week for our exam this Friday. However, considering the circumstances that happened this week Mrs. Robinson changed the exam to just chapter five. Chapter five is about exponents. We covered a few sections in five and it is pretty easy stuff. We learn how to solve exponents and we used 3 different formulas. Let’s do some example problems.
Solve the following problems:

Example 1:
7-2 = 1/49

Example 2:
(-7)-2 = -1/49

Example 3:
When given a chart you must pick one of the following formulas:
Read the problem, plug in the numbers, and then into your calculator and get an answer.
A(t) = Ao (1+r)2
A(t) = Ao b t/k
P(t) = Po ert

Rational Exponents

This week in advanced math I think it is chapter 5 which we are working on. One of the sections we are doing is called rational exponents. This is not to difficult I will show you how to do some of these kinds of problems.

FORMULAS:
A(t)=A0b^(t/k)
A0 is starting point
b is halfing, doubling, ect
t is time and k is time it takes to get to b

Exponents that involve roots.

EXAMPLE:
5th root of (3)^3
3^3/5

(4/16)^1/2
2/4 which is equal to 1/2

*This is the easiest way to explain rational exponents.

this past week was the worst week of school ever. we learned about rational exponents. these are very easy and we've seen this in aleks. everything we did this week was referred to aleks. so here are a few examples of stuff.

EXAMPLE:
6square root of x^3--------x^3/6

FORMULAS:
a(t)=a0(1+r)^t
0---is your starting point
r---is your decimal
t---is your time

*when given a word problems you can easily plug it into the formula. once you have it plugged into your formula you can then put it in your calculator to get your final answer. It's really simple so i hope we have a good bit of these on our 9 weeks exam friday :)

i've still been trying to do aleks but it keeps kicking me off my computer. i am learning with using aleks.

Exponents

This was a hectic week in school and advanced math. We learned about all kinds of exponents and how to change them from exponents to radicals and vise versa. We did a lot in Aleks, and we had a lot of homework this week. We also learned about rational exponents, which are exponents that involve roots. This was pretty easy with just the numbers. We we inserted letters it took me a little while to get the hang of it, but I figured it out. I enjoyed Rational Exponents better than the T1 formula so here you go.

Rational Exponents

Ex. (49/81)^1/2

1) (49^1/2)/(81^1/2)

2)(square root of 49/ square root of 81)

3)7/9

5-4
Almost forgot about this! This week in Advanced Math we learned about exponents and formulas involving percent, rate, and time. Section 5-4 involved one formula – P(t)=Pe^r*t
P – starting amount
e – on calculator (e is approximately 2.7)
r – rate as a decimal
t – time

EXAMPLE 1: Suppose $2,000 is invested at 5% interest compounded continuously. How much money do you have after 3 years?
A). Using the formula, plug in the information given.
P=2,000(starting amount)e^(0.5*3)
B). By plugging all of that into your calculator, you get $8963.38 as your final answer.

EXAMPLE 2: Which has a larger value? e^square root of 3 or 3^e
A). The answer to this problem can be achieved by plugging the two into your calculator.
B). e^square root of 3 = 5.65 and 3^e = 19.81
C). Therefore, the first one has a larger value.

This is a little short but my computer is installing new updates and it’s counting down to shutdown!

This was a wild week in math and school in general. From matrices to carnival ball, we experienced it all this week. Now it is time for my least favorite part of the night. The part where I have to blog late at night, and come to the realization that the weekend is over and school is tomorrow. But anyways heres your blog.

Matrices

This week in advanced math we learned how to do matrices. It is a long tedious process but it isnt too hard.

Ex.

x y z

2 3 4

3 2 1

x[3*1 + 4*2] + y[2*1 + 3*4] + z[2*2+3*3] = x[11]+y[14]+[13]= 38

blog.blog.blog.

hellooo everyonee:)
well this week everyone has really been busy with the carnival ball and what not. But despite all the hectic over that i must admit that the week was pretty productive for my advanced math and physics classes anyway..but anywho we learned a lot vectors again. When we were'nt learnning about vectors, we were working on our ALEKS which is gonna be done pretty soon.

we learned how to tell if vectors were parallel or perpendicular..you can do this by doing dot product. if the answer is equal to zero, then vectors are perpendicular.

ex. ( 2,3) (0,0)
(2*0)+(0*3)
=0+0
=0
therefore, the two vectors are perpendicular :)

in order for two vectors to be parallel they must be proportional. Here is an example of two vectors that are proportional, and parallel.

ex. x(1,2) y(2,3) z(3,4)

2/1=3/2=4/3
2=3/2=4/3
so noo, these vectors are not parallel which is too bad :(

Alright this week we were really crazy. There was carnival ball and stuff. We took a test on Thursday and Friday we didn’t have math class and a bad thing is I forgot my whole binder. If I do it on something easy I’m sorry and hope you forgive me.

Examples:

Matrices:

2[ 1 2 4] + 3{3 2 5]

{2 4 8} + { 9 6 15}= [ 11 10 23}


Named the matrices

{ 1 3 4 5}
{ 2 4 8 9} = 2x4


find AB
A(1 4 8) B(5 4 7)= (4, 0, -1)


point 8,-2, 3 and 4, 0, 7

Find center = (6,-1,5)

Radius square root (4-8)^2+(0+2)^2(7-3)^2=16+4+16 = square root of 36= 6

Equation = (x-6)^2+(y+1)^2+(2-5)^2=9^2

This week we learned a few new sections, had to complete a total of 70 modules on aleks, and took our Chapter 12 test. The next week our whole blue pie will have to be complete. In the section below I will explain our notes on 3D. This Chapter is mostly formulas and is pretty easy to remember if you study it.

Formulas you need to know for this section are below:

AB = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2

Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)

Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2

Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)

Vector addition, magnitude, dot product, etc.are just a little different but follow the same formulas as earlier sections. They just have an extra variable in the formula unlike before.

Example

Find the center and radius Point A (8,4,2) Point B (2, 8,6)

Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)

Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 6+ 6 + 6= square root 16/2 = square root 8

Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 8

Alright soooo, im glad carnival ball is over. But now i'm going to fail this class. ALEKS is not my friend anymore. I work too slow and i havent met the latest deadline which will cause my grade to go wayyyyyyy down. The assessment test sucksssssssssssss.
glad ive said that now on to some math.

Finding whether something is parallel or perpendicular is my favorite.

EXAMPLE:
j=(3,4,2) k=(15,8,4)
PARALLEL:
15/3=5
8/4=2
4/2=2

PERPENDICULAR:
3(15)+4(8)-2(4)
**in order for it to be perpendicular it must equal 0.
****therefore its neither.


The AB thing is easy too.
A(2,3,1) B(4,2,6)
4-2=2
2-3=-1
6-1=5
*your answer is (2,-1,5)

my computer keeps freezing so im going to post now.

Hey guys so this blog is going to suck this week because I have had a super, busy, exhausting weekend with carnival ball, and we had a field trip and we did not have class on Friday, and I forgot to bring my notebook home so I am going to have to go off of some super easy stuff I remember so forgive me. Let’s see matrices are pretty easy so let’s just do some random examples cause I did awesome on my on this entire chapter! Yay!

Random example of matrices:

Solve the matrices:

2 [ 3 4 5 ] = [ 6 8 10 ]

[ 7 8 ] + [ 9 10 ] = [ 16 18 ]

Name the matrices:

[ 2 4 5 6 9 ] is a 1 x 5

[ 9 10 15 ]
[ 1 12 20 ] is a 2 x 3

Vectors

This week in advanced math we have just finished chapter 12. Chapter 12 was about vectors. We took a test on vectors and I thought that the test was fairly easy. The hardest part was the 4x4 matrix which wasnt really that hard; just time consuming. Today I am going to teach you how to add and subtract a vector. This is probably one of the easiest things you will learn to do.

DIRECTIONS:
1. You should make sure you have a vector, u and v are always vectors.
2. Then you should add x1 to x2 and you will come out with a new vector.
3. You should do the same thing for subtraction just simply subtract the vectors.

EXAMPLES:
u=<4,5> v=<3,4>

Addition:
Find u+v
<4,5>+<3,4>
answer:<7,9>

Subtraction:
Find u-v
<4,5>-<3,4>
answer:<1,1>

*This is the easiest thing you will learn to do with vectors from my point of view.
*Just remember add the x's and add the y's and the same rules apply for subtraction.

3x3

This past week we learned how to do 3x3's and 4x4's. I will show examples of the 3x3 because the 4x4's take FOREVER and so much paper. However, they are pretty much the same concept so if you are able to do one, you can do the other. 3x3's are very simple once you are taught the correct way because it is all simple math. So, here we go:

Example 1: (I won't have the brackets around them because I don't know how to do that)

6 2 3
1 2 4
0 1 1

1. you want to pick the row with the smallest numbers and cross it out.
2. cross out the first column, so you are left with:

0 * 2 3
2 4
3. Now cross out the second column, and so on ( alternate the sigsn of subtraction and addition.)

0 * 2 3 + 1 * 6 3 - 1 * 6 2
2 4 1 4 1 2

4. Now just multiply your matrices.

8-5= 3 *0 = 0
24-3 = 21 * 1 = 21
12-2= 10 * 1 = 10

0 +21 -10= 11

11 is your answer!

3-D
The formula to find midpoint is (x1+x2/2, y1+y2/2, z1+z2/2)
IABI = square root (x2-x1)squared +(y2-y1)squared + (z2-z1)squared
Vector equation = (x, y, z) = (xo, yo, zo) + t(a, b, c) *a, b, and c is your vector

EXAMPLE 1: A sphere has points (6, 2, 3) and (2, 2, 7). Find the center and radius.
A). Use the midpoint formula first.
(6+2/2, 2+2/2, 3+7/2) = (4, 2, 5)
B). Now use the IABI formula.
Square root (2-6)squared + (2-2)squared + (7-3)squared = 0
C). The answer you get there is your diameter.
D. Now plug the answer you got from step A into the vector equation formula.
(x-4)squared + (y-2)squared + (z-5)squared = 0 (radius squared)

EXAMPE 2: Find the vector and parametric equations of the line containing (1, 3, 5) and (2, 8, 9)
A). Use your vector equation formula.
(x, y, z) = (1, 3, 5) + t(1, 5, 4)
*(1, 5, 4) is found by subtracting x2-x1, y2-y1, and z2-z1
B). x = 1 + t, y = 3 + 5t, and z = 5 + 4t


Alrighttttttttttttttt, hope everyone has a great weekend and good luck to everyone at Carnival Ball :)

Inverse for a 3x3

This week in advanced math we are working on chapter 12. Mrs. Robinson currently showed us how to find the inverse of a 3x3 matrix. This was fairly easy for me because I was already taught it in the number theory class that I am taking, because of this Mrs. Robisnon let me get a head start on my homework (HA HA HA). Anyways I am going to teach you how to do this.

First, you are going to have to pick a row with the lowest numbers in them, then delete the row in column that they are in leaving you with a 2x2 finding the determine of that then multiplying the outside number by the determinate roatating signs postive and negative, starting with positive, and then add all the determinates up.

EXAMPLE:
1. [1 2 2
0 0 1
1 3 0]
0[2 2 0[1 2 1[0 0
3 0] - 1 0]+ 1 3]
0-0+0= 0
so your answer for this one happens to be 0.

3-D

This week we learn a lot of matrices and vectors. But this section is 3-D aka the hardest section.

|AB| = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2

Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)

Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2

Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)

Vector addition, magnitude, dot product, etc. all follow the same formula with extra variable.

Example: point 8,-2, 3 and 4, 0, 7

Find center = (6,-1,5)

Radius square root (4-8)^2+(0+2)^2(7-3)^2=16+4+16 = square root of 36= 6

Equation = (x-6)^2+(y+1)^2+(2-5)^2=9^2


Basically if you know your formulas you will do pretty good at this section if not you will do horrible at this section.

This week in advanced math we started on chapter 12. We also delved deeper into the Aleks program. I absolutely hate that program because Im the type of person with little time, I dont have all day to sit at a computer because Im never home. In other words I miss blogs so bring them back and get rid of Aleks. But anyway back to chapter 12.

Vectors and Planes

-ax + by + cz+ d when ax + by + cz= d

|a,b,c| must be a vector perpendicular to the plane

3D graphing

-Perpendicular vector to a plane are said to be normal.

Ex. idk if this is right

Vector( 6, 5, -4) is perpendicular to a plane that contains A(0,1,2) Find an equation of the plane.

4x-5y+6x=24

This week we focused more on lessons in class and less on alex. We are finally get back into trig and preparing for our trig exam. Next week our blue pie for alex will be due so we can get into logs and more things with trig. In the section below I will explain our notes on 3D. Chapter 12 has been a lot of formulas and has been pretty simple if you study.

Formulas you need to know for this section are below:

AB = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2

Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)

Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2

Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)

Vector addition, magnitude, dot product, etc.are just a little different but follow the same formulas as earlier sections. They just have an extra variable in the formula unlike before.

Example

Find the center and radius Point A (8,4,2) Point B (2, 8,6)

Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)

Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 6+ 6 + 6= square root 16/2 = square root 8

Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 8

Math 3 dimensions

School is getting old B-Rob. We need to change school to be like 3 days a week, but thanks for not giving us bookoo homework. This week we did some stuff with vectors, except these had a third dimension. Everything we do for them is exactly the same, except you just add an extra set of numbers.
The formula used is- square root (x2-x1)^2+(y2-y1)^2+(z2-z1)^2

Ex: Find |AB| for the points ( 2,3,2)  (6, 7, 4)
Square root (6-2) ^2 + (7 – 3) ^2 + (4 – 2)^2
16 + 16 +4 = 36
Square root 36 = 4; |AB| = 6

3D

This week we did a lot of different things with vectors and matrices. They all seemed pretty hard when they were taught but they are actually not that bad and for the first time I was able to fly through one of the homeworks. Here are the 3D notes, which was probably the hardest section for me:

For magnitude: |AB| = squareroot (x2 – x1)^2 + (y2 – y1)^2 + (z2 + z1)^2

Ex: Find |AB| for the points ( 2, 3), (3, 7), (1, 0)
Square root (3-2) ^2 + (7 – 3) ^2 + (0 – 1)^2
1 + 16 -1 = 16
Square root 16 = 4; |AB| = 4

Ex2: Find |AB| for the points (3, 6) , (3, 0 ) , ( 4, 7)
Square root ( 6 – 3 ) ^2 + ( 0 - 3 ) ^2 + ( 7 – 4 ) ^2
9 -9 + 9 = 9
Square root 9 = 3; |AB| = 3

Ok so this week has been kind of a blur. For some reason I can’t remember exactly what we did this week except I know we learned a couple of new sections. One of these sections has something about 3D. I can’t even remember the section number but anyway it has something to do with vectors and midpoint equations. Here are the notes Mrs. Robinson gave us.

|AB| = square root (x2-x1)2 + (y2-y1)2 + (z2+z1)2

Midpoint = (x1 + x2/2) (y1 + y2/2) (z1 + z2/2)

Equation of a Sphere = (x-xo)2 + (y-yo)2 + (z-zo)2 = r2

Vector equation (x,y,z) = (xo, yo, zo) + t(a,b,c)

Vector addition, magnitude, dot product, etc. all follow the same formula with extra variable.

Example

Find the center and radius Point A (9,5,7) Point B (7, 3,5)

Center – 9 + 7/2 = 8, 5 + 3/2 = 4, 7 + 5/2 = 6 (8,4,6)

Radius – square root (7 - 9)2 + (3 - 5)2 + (5 – 7)2 = 4 + 4 + 4 = square root 12/2 = square root 3

Find the equation
(x – 8)2 + (y – 4)2 + (z – 6)2 = 3

12-1

12-1
Section 12-1 was about vectors. Vectors represent motion and have a magnitude and direction. To add vectors, you add head to tail. To add vectors algebraically, you add components. To find the magnitude of a vector, you use the formula square root of x^2 + y^2. When given two points, such as A and B, and asked to find the vector, you use the formula (x2 – x1, y2 – y1).
Having these few little notes, let’s try an example.

EXAMPLE 1: Given A(6,4) and B(8, 9), find the vector AB.
A). According to the notes above, we’re going to use the last formula.
B). First, subtract the x’s (8-6) = 2
C). Now, subtract the y’s (9-4) = 5
D). Your vector is (2,5)

EXAMPLE 2: Find the magnitude of EXAMPLE 1’S answer. **MAGNITUDE HAS TO BE WITH A VECTOR!
A). Since we’re finding the magnitude, use the square root formula.
B). square root 2^2+5^2 = square root 4 + 25
C). Your magnitude is square root 29

Advanced Math

This week in advanced math we learned matrices and we also started to learn chapter 12. Chapter 12 has to do with vectors.

• Vectors represent motion
• Vectors have a magnitude and a direction
• To add vectors you add head to tail.
• The triangle formed by the vectors give the resultant vector
• To add vectors algebraically you add components

To find the magnitude of a vector you do (x2-x1, y2-y1)

Ex. A(11,6) B(12,7) Find vertices AB

(7-6),(12-11)

AB=1,1

Chapter 12

This following week we got back into trig. It was weird and kinda hard to switch back after having such easy lessons the past weeks before. We were introduced to Chapter 12 and also did a lot of work on aleks. It was a hectic week with all the "snow" threats and the rainy weather. I understand Chapter 12-Section 1 more than anything else because there wasn't much to it. It was mostly just having to connect the vectors, add on to them, or flip them around.


vector- is a line that represents motion

• line has a magnitude and a direction, which are the points that are given.
• ( x is the head and y is the tail )
• when is given, it is not asking for an absolute value like usual. that symbol means find the magnitude.

When working with a vector, it usually ask for 3 things.

Example:

Find 4t

• That is telling you to take the vector that it gives you and draw it four times connecting in a straight line.
• If it asks you to find –t, you draw the vector facing the opposite direction.
• Lastly, if it wants you to find t + q, you connect the two vectors together, usually creating an angle.

This past week we did a lot of stuff with aleks. It is really helping me alot. If you do not understand something all you have to do is click explain and it explains it to you. It has helped me a lot with like, simple math and stuff. The only thing i do not like is that it kind of like stalks you because it can tell if you click on another tab and stuff. And it is soooo many problems. But other than that, it is pretty helpful. This week we also learned about vectors. They represent motion. If you are adding vectors you add them from head to tail, but if you are adding them algebraically, you add the components. The triangle that you get after you add the vectors from head to tail is the result of what you added. They have a magnitude and a direction. I can not relaly do any examples because its all drawing graphs and such.

..... Vectors?

All of my days are combining together so, I'm not entirely sure if this is from this week or the week I missed. But here we gooo....

Vectors represent motion. They have a magnitude and direction. They are written with a --> over the two given points.

To add vectors, you must add head to tail. The triangle formed by the vectors give the resultant vector. This basically means that the imaginary line you get when you have
^ \ is the answer.
/ v

To add vectors algebraically, you have to add the components. To do that use this formula: (x (subscript 2)-x (subscript 1), y(subscript 2) - y(subscript 1))
-->
Example-- Find AB A(4,2) B(9,-1)
(9-4, -1-2)
=(5,-3)

To find the magnitude, plug the answer you get from the algebraic vector into this formula: square root(x^2 + y^2). Magnitude is shown as | |
-->
Example-- Find the Magnitude of AB
Square root(5^2 + -3^2)
Square root(25+9)
=Square root(34)

chapter 12

This week was really frustrating because we are getting back into trig after having some pretty easy lessons lately. We covered sections 12-1 and 12-2.

12-1 was pretty easy because all there really was for us to do was connect the vectors, add on to them, or flip them around. A vector is a line that represents motion. The line has a magnitude and a direction, which are the points that are given. ( x is the head and y is the tail ) also, when | | is given, it is not asking for an absolute value like usual. | | that symbol means find the magnitude.

As for the vectors, there are 3 things that it usually asks:

For example: Find 4t
That is telling you to take the vector that it gives you and draw it four times connecting in a straight line.

If it asks you to find –t, you draw the vector facing the opposite direction.

Lastly, if it wants you to find t + q, you connect the two vectors together, usually creating an angle.

I think we should continue to go to the library sometimes to do aleks. I find its really helpful and not as boring as doing work out of the book. Sorry this was such a short blog.

Math

This week we did Aleks a lot. Aleks is great. Every time I get a problem right it tells me how great I am and its a good motivater. We need to spend more CLASS time doing Aleks, not just  home time. It wouldn't be so bad for at home but 10 units is a lot B-Rob. I don't really remember what we did to much this week so i'll just put my notes from section 12-1 up hear. BTW we had baseball on Superbowl Sunday which is ridiculous.
 vectors represent motion
they have a magnitude and direction.
to add vectors you add head to tail.
the triangle formed by the vectors is your result.
to add vectors algebraically you add the components.

This week we continued to work on aleks.com. I love aleks because i can work at my own pace and i'm learning a lot. We finished up chapter 14 on matrices (which i absolutely love). Now we are working on vectors. chapter 12-1 and 12-2.

12-1

VECTORS-represent motion
they have a magnitude and direction.
-to add vectors you add head to tail.
-the triangle formed by the vectors is your result.
-to add vectors algebraically you add the components.

12-2

-to find the magnitude
**square root of x^2+y^2

-scalar multiplication puts a scalar times each component.
-you are given 2 points (A&B)
**A&B=(x2-x1;y2-y1)
^^^^remember this from algebra 2!

*it's hard to do examples for 12-1 since they are graphs drawn.

Now im going to spread my love for aleks.com
it is a whole lot easier than doing 2 blogs per week.
im learning from the stupid mistakes i usually do.
its fun?
*i think as a class we all enjoy it. (especially when we do that in class)

byeeeeeeeeeeeeeeee.

14-3

This week in advanced math we finished up chapter 14 and took a test and now we are currently working on chapter 12. Chapter 12 is about vectors but I am going to show you how to take the inverse of a matrix today. This is probably one of the easiest things to do when it comes to matricies. Here is how you find an inverse of a matrix.

1. First you are going to want to switch the first number of the matrix with the fourth number.
2. Then make the second and third numbers opposite of their sign.
3. Then cross multiply your origional matrix and find the determinate.
4. Take your determinate and multiply it by your second matrix.

EXAMPLE:
1. [2 3
1 4]

[4 -3
-1 2]

(4x2-1x3)= 5

1/5[4 -3
-1 2]

[4/5 -3/5
-1/5 2/5] This is your answer.

This is the easiest way to do this just follow the steps and you should be fine.

12-1,12-2

This week we learned some sections in chapter 12. In 12-1 and 12-2 we learned about vectors and how to represent it.
Vectors-represent motion and it has a magnitude and a direction.

*to add vectors you add head to tail.

the triangle formed by the vectors give the resultant vector.

another thing is to add vector algebraically you add components.

-to find the magnitude

square root of x^2+y^2 magnitude is done as |->|

Ex: A(4,2) B(9,-1) find on top of AB is ->
on top of AB ->(9-4,-1-2)=(5,-3)is on top of AB ->

find the magnitude of on top of AB ->

square root of 5^2+-3^2= square root of 34

Ex: find |on top of AB ->| if A(1,3) B(2,4)
on top of AB -> = 2-1,4-3= (1,1)

|on top of AB ->| Square root of 1^2+1^2= square root of 2.

If you know all your rules and stuff sections 12-1 and 12-2 will be easy for u.

Sections 12-1 and 12-2
This week in advanced math we took our test on Wednesday instead of Tuesday because we got out of school early due to the weather. The test was super easy so hopefully I got a good grade. Since we took the test on Wednesday we had to double up on learning sections in chapter 12. Sections 12-1 and 12-2 are on vectors. Vectors represent motion. They have a magnitude and a direction.
To add vectors you add the head and the tail. To add vectors algebracically you add the components.
The triangle formed by the vectors give the resultant vector.
To find the magnitude square root x2 + y2 magnitude is done as | -> |
Scalar multiplication puts a scalar times each component.
Given 2 points A and B
->
AB = (x2 – x1 y2 –y1)

Geometric Representation

Find c + d
Connect the head of C to the tail of D.

Find 2s
Connect the head of S to another tail of S.

Find –p
Reverse the p and draw it in the opposite direction.

14-3

Here's my weekend bloggggg:
This week we finished Chapter 14 and tested on it. Along with that we continued to work on Aleks and even started Chapter 12 which had to do with vectors. I’m going to explain the concept of finding an inverse in a matrix.
EXAMPLE 1: Given A= [4 3 find its inverse.
2 6]
A). To find an inverse, flip the 1st number on the top row and the 2nd on the bottom (4 and 6) and make the 2nd number on the first row and 1st on the second row negative (-3 and -2).
B). Your new matrix should look like this [6 -3
-2 4]
C). Now cross multiply your original matrix and subtract (4 x 6 – 2 x 3) = 18
D). Mutiply your new matrix by the determinant, which is 1/18.
E). Your answer is [1/3 -1/6
-1/9 2/9]

I know this is a little short but I've been doing Aleks all morning and I'm really frustrated :/ sorry!

alkjvhsak;lg ;lsg

aleks.com

This week we were introduced to Aleks.com which is a website to help better our math skills. I have been working on a lot of old algebra things I have lost my grip on and I like that the website is helping me get back into the hang of the old concepts. So for my blog I am going to go back to Algebra and go over solving a linear equation.

Im just going to do a few examples because im not sure how to explain this concept to well.

Example:
-2= 7x-9/6- 8x+5/7
First we are going to examine the problem and determine the least common demominator.
The least common denominator is 42.
Multiply 42 by each segment of the problem.

(42)-2= 42(7x-9/6-8x+5/7)
(42)-2+ 42(7x-9/6)-42(8x+5/7)
Now divide into the LCD.
-84= 7(7x-9)-6(8x+5)
-84= 49x-63-48x-30
-84=x-93
Now solve for x and
answer is.... x=9

Matrices are AWESOME

Yeah thats right read the title. It is because of these row by column organized numbers that I will be able to keep my grades up for now. They are simple and easy to do and having something easy to do could not have come at a better time. Baseball is usually a pretty laid-back sport where you go have fun, but since the season has yet to start it has become a boot camp. I am physically and mentally exausted from practicing everday and having a break from hard math is only helping. Matrices are easy to add and subtract. If they are of the same dimensions you simply add or subtract them with their corrresponding number. If not you can't do anything. Multiplication is a little more complicated but not really. The columns of the first matricee has to be equal to the number of rows in the second. Then you take the number of rows in the first and combine it with the number columns with the second and get the dimensions of the product. After all of this you do some fancy math and get the answer. This blog was fun. Not Really. I'm goin to sleep. Night.

Matrices

 

    Well helloo :) my weekend has been going well so far in case you all were wondering. But anyways, this past week we learned about Matrices and we also began the aleks program on the computer. Other than that we pretty much reviewed because we had already learned about matrices in algebra 2. Matrices are pretty simple to work or solve but they are a bit tedious.

There are some rules for matrices:

When adding matrices, they must be of the same dimensions. The same goes for subtracting matrices.

Ex. A 2x3 matrices can only be added to a 2x3 matrices

When multiplying matrices, you can only multiply matrices that have the same colomns and rows.

Ex. 2x3* 3x1 the colomn of the first are equal to the rows of the second.

You cannot technically divide a matrices, to "divide" a matrices, you simply multiply by the inverse..

Ex. inverse

0 1

1 0 = 0-1=-1 = 1/-1* 0 1

1 0 = 0 -1

-1 0

This past week we learned matrices and worked on aleks. Matrices are super easy and stuff.

ADDITION:
[2 6]+[5 4]=[7 10]

SUBTRACTION:
[9 2]-[7 1]=[2 1]

MULTIPLICATION:
5[2 4]=[10 20]

So i'm really tired and I probably shouldn't have waited until the last minute to do my blog but oooops. I guess it will just have to be a short one but next week i'll make it longer :)

14-1

Section 14-1
This week in advanced math we started chapter 14. Chapter 14 is all about matrices. Matrices are super easy stuff we learned in algebra 2. We finished up our chapter 4 test on Monday and took a quiz either Tuesday or Thursday on matrices but I can’t remember which day. We also started with this new online math website that is kinda like skills tutor. It’s pretty cool I guess, helped me remember some stuff I forgot from algebra 2. Anyway let’s do some super easy problems on matrices.
Examples:
5 [3 6] = [15 30]
[9 10] [45 50]
*this is supposed to be a 2x2 but idk how to do big brackets so all my examples will look pretty
wack so deal with it.

[1 3] + [2 4] = [3 7]
[5 7] [6 8] [11 15]

Name the matrice

[5 6 7]
[4 1 -3]
[3 -1 -4]

3x3

Matrices

This week we learned how to do matrices. You can add, subtract, multiply, divide, square, cube... ect... matrices. They have other term that we learned to like transpose and trace. But today I am going to teach you how to add, subtract, and transpose matrices.

ADD:
1. You are going to add the first number in the first matrix to the first number in the second matrix.. you do this for the second, third, forth, and so on.. This rule applies the same for subtracting.
Example:
[1 2 [1 2 [2 4
3 4] + 3 4] = 6 8]
This is as easy as it gets.

2. Subtracting- do the same as adding but you subtract!
Example:
[1 2 [1 2 [0 0
3 4]- 3 4] = 0 0]
This is also as easy as it gets.

3. Transpose- when transposing you just make the row a column.
Example:
[1 2 3 4
5 6 7 8
9 9 9 9]

now transpose it.
[1 5 9
2 6 9
3 7 9
4 8 9]

** This are the easiest possible ways to add subtract and transpose matrices.
I hope you all got the hang of it, if not too bad cause you will officially be called an idiot.

Matrices

This week we learned 14-1 and 14-2. In 14-1 and 14-2 is it all about matrices and how to add, subtract and multiply and determine if they are matrices.

Row and column (row go across and column go up and down)

Examples:
1.
-3{1 4} = {3 12 }
9 6 27 18

2.
-{ 1 1 1] determine row x column= 4x3
1 1 1
0 0 0
2 2 2

3.

A={ 2 6}
8 22 a find a^t
9 16

B. A^t= {2 6}
8 22
9 16

C. { 1 2 }{ 2 4}
4 6 0 1

2x2 2x2

Middle two are the same with you do now
{ 2+0 4+2}
8+0 16+6 ={ 2 6}
8 22


{ 1 2} + {2 0} ={ 3 2}
4 6 5 2 9 8


Basically if u remember everything from last year in algebra 2 it is the same thing and all this is review from last year just know your rule about matrices and you will be fine. Btw my brackets sucked I tried my best making it but oh well hope u like it.

♥ Marjorie Ann Flanagan St. Martin ♥

This week I did not go to school except for one day, but I didn't stay the whole day and missed 7th hour. But, I did learn lessons this week. I learned how wrong doctors can be. I learned how truly idiotic River Parish Hospital is. I learned how sweet Oschner nurses are. I learned you can survive with one lung and never know the difference. I learned how strong a family can be. I learned how powerful one woman can be. I watched an amazing woman tell a doctor thank you and pat his hand consolingly after he told her she was going to die. I watched this same woman peacefully and gracefully take her last breaths.
Gram was the classiest, sweetest, most wonderful woman to walk this earth. While I miss her dearly and mourn over her loss, I think of how great her life was and how she'll be with her husband again. I think of how happy she must be because there is no doubt about her being in heaven. I love my grandma so much. It's hard to believe she's gone.

Rest In Peace Marjorie Ann Flanagan St. Martin. <3

Matrices

Okay so this week we learned matrices and started the aleks stuff on the computer. The first two sections of matrices were easy for me because I remember them from last year. Its going to be super hard to type out matrices because I don’t know how you can put more than one row in a bracket. :/

The basic part is finding out how big your matrices is: rows x colums
(rows go across and columns are up and down)

Ex 1: [ 2 4 6]
This matrix is a 1x3

To add matrices is very simple. You just add the corresponding numbers.

Ex 2: [ 4 2 4] + [ 7 3 6] = [ 11 5 10]

Super easy and subtraction is done the same way.

Ex 3: [ 8 3 2] – [ 5 4 1] = [ 3 -1 1]

Sorry all of my rows were 1, but I have no clue how else to do it!

Matrices!

Woaaaah, feels like I haven’t done one of these in forever!
We started Chapter 14 this week and it’s all about MATRICES! I’m not really sure how I’m going to type them out so bare with me. You can add and subtract matrices but they have to be the same dimensions. Another thing we learned was how to multiply them. When doing so, you have to have the same number of columns in the 1st one and rows in the 2nd one. We also learned how to find its inverse.

Rows – go across
Columns – up and down

Adding and subtracting matrices is simple..
EXAMPLE 1: [-5 0 + [6 -3
4 1] 2 3]
A). To add matrices, we add the corresponding members. Subtraction is done in the same way.
B). (-5+6) (0-3)
(4+2) (1+3)
C). [1 -3
6 4]
D). Your new matrix is a 2 x 2.

EXAMPLE 2: 3[-3 0
4 5]
A). Multiply each number of the matrix by 3, easy.
B). [-9 0
12 15]

When adding, subtracting, and multiplying matrices be careful of negatives! This blog is nasty looking because of those stupid brackets :/

4-5

Last week we learned chapter 4 section 5. in 4-5 we learned about Inverses and how to determine the horizontal line test.

Horizontal line test- similar to the vertical line test only if it passes then there is an inverse that is a function

Sidenote-for our book they will say there isn’t an inverse if it fails

To find an inverse:

-Switch x & y

- solve for y

To check if something is an inverse:
F(g(x))=x
&
g(f(x))=x

-f^-1= inverse notation

Examples:

Prove f(x)=x+3 f^-1(x)=-x-3 are inverse

F(f^-1(X))=x-3+3=x

F^-1(f(X))
(x+3)-3=x

yes it is an inverse.

Y=square root of x-3

x=square root of y-3

square both side you get y-3=x^2
y^-1=x^2+3

it is an inverse because it passes the horizontal line test.


Basically if you know all these formulas and stuff section 4-5 will be easy for you.

Inverses Chapter 4

This week we continued Chapter 4. We were introduced to inverse functions in Section 5. This chapter was fairly simple being the only thing you had to do is switch the x’s and y’s in the equation and then simply solve for y. For this section there weren't many steps to the process but before you did anything to the function you must graph the equation on a coordinate plane. Once you have the equation graphed you use a method called the horizontal line test and test the function. If it passes and is an inverse you can move alont the rest of the process.

How to do a horizontal line test: 1.draw horizontal lines across the graph.
2. if it doesn't touch more than one point you can find the inverse.
* Note that in college we will be taught differently and they'll show you a way to find the inverse even if it fails the test.

To show something is an inverse be sure to put the -1 exponent after the y once you have solved the equation.

Example:Find the inverse for y = x -9
You graph this equation, and do the horizontal line test, and it does pass.
1. Switch the x and y. x= y-9
2. Solve for y. y = x + 9
3. Y^-1 = x + 9

Chapter 4

This week in advanced math we continued in chapter 4. We learned how to decide if a graph was a function and now we learned how to tell if a graph is an inverse. You do this by the vertical line test. We also learned how to find an inverse.

To find and inverse:

1) switch x and y

2) solve for y

If the formula is squared or or taking the absolute value then it is not an inverse.

To check if something is an inverse:

1) do f(g(x))= x and g(f(x))

Also in section 4-4 we learned about stretching and shrinking graphs.

• When you multiply a variable by a number less than 1 it makes the graph shorter and fatter.
• When you multiply a variable by a number bigger than 1 it makes it skinnier and longer.

ALMOST 17 BABY!!!!!!!

That right my b-day is tomorrow. I'm going to be 17 years young. That means i"m old enough to finally get my liscence and drive. How cool is that? Mrs. B-Rob I think you should give me some points as a present. Anywho, we did Math this week. I'm super proud of myself because I believe I did good on most of the tests we were given this week. I can't particularly remember if we learned anything new or not. It's been a pretty long weekend. We did learn some stuff recently about symmetry lately though. This is pretty simple. You have to find if an equations graph is symmetrica on the y-axis, x-axis, the origin, and as well as if y=x.
You may use these steps to perform the process.
x-axis. Step1. Make all y's in the equation negative.
            Step2. Simplify the equation. If the equation is the same then it is symmetrical.
y-axis. Step1. Make all x's in the equation negative.
            Step2."......"
origin.  Step1. Make both the y and x's negative.
            Step2. "......"
y=x       Step1. Switch the x and ys
            Step2. solve for y
            Step3. If equations are equal than it is symmetrical.

4-5 Inverses
This week in advanced math we reviewed for a chapter 4 test. We took a took quizzes on Tuesday and Thursday but the quiz on Tuesday didn’t count. Thank you Mrs. Robinson! We took part of the test Friday and we will take the other test on Monday. Section five of chapter four is on inverses. Here is a few notes of inverses:
Horizontal Line Test – similar to the vertical line test only if it passes then there is an inverse that is a function.
To find an inverse – 1. Switch x and y 2. Solve for y
To check if something is an inverse f(g(x)) = x and g(f(x))
f-1 – inverse notation
Side note – In our textbooks it says if there isn’t an inverse if it fails.
Example:
y = x – 7
x = y – 7
y-1 = x + 7
When you draw the graph it passes the horizontal line test.

Chapter 4

In chapter four we learned about functions and how to find their inverses, symmetry, and if they were functions at all.

Finding out if a function is a function at all involves the use of the vertical line test.

When you graph a function; if you can draw a vertical line and have it hit two points at any part of the graph, it's not a function.

The function f(x)=x^2 + 4 would be a function since there's no point on the graph where a vertical line would hit a x point twice.

The graph x=2 is not a function because a vertical line hits all the points on the graph.

Finding a function's inverse

If a function has an inverse, a horizontal line will not be able to touch any two points of the graph.

The graph f(x)= x+3 has an inverse because a horizontal line would cross its graph once.

The graph f(x)=x^2-5 wouldn't have an inverse that is also a function because a horizontal line can touch the graph twice.

Chapter 4

This week in advanced math we learned chapter 4. This was probably one of the easiest chapters that we have done all year. Well Friday we took the multiple choice portion of the chapter test and now Monday we will be taking the free response portion of the test. I think that I did really well on the multiple choice portion of the test. Hopefully the free response test will be just as easy. Now lets talk about INVERSES.

First: You should graph the equation and make sure it passes the horizontal line test.
(this is so you do not waste time finding an inverse of something that isn't mathematically possible.)
Next: If an inverse, you should switch the x and y and solve for y.

**HINTS:
*If it is x^2 it wont have an inverse because it will be a parabola. (no inverse)
*If it is x it will be a straight line (has an inverse)
* If it is x^3 it will make this weird curve line thingy (has an inverse)

EXAMPLE:
1. y=x^2+4
answer: no inverse (this is a parabola)

2. y=x-4
x=y-4
y^1=x+4 is your answer

I hope that I explained this well and if you think I didn't then you are wrong because I did. Thanks for taking up 5 minutes of my time for 5 extra points. :)

4-5

This week was weird. I really don't remember too much of what went on in school. Friday, we took a test and the other days were review, I believe. One of the sections on the test was section 4-5. Section 4-5 was about finding inverses.

To find an inverse you have to use the horizontal line test. This is like the vertical line test, but it's horizontal. If it passes without touching two points on the graph then there is an inverse that is a function.

This brings us to the first step - drawing the graph. Since I don't have a fancy laptop, I can't draw a graph. But, here are some hints to see if it will pass the test. If it has an x^2, it will not pass the test because x^2 makes a parabola. An x makes a straight kinda slanted line.

After you graph, to find an inverse switch x and y. Then solve for y.


For example, y = x + 3

First, graph the equation.

By using the horizontal line test, you know that there is an inverse.

Now, switch the x and y.

x = y + 3

And now, solve for y.

y^-1 = x - 3

Place -1 above y to show that it is the inverse.

So this past week we went over a few things from the week before for our chapter test and we learned some new things. Section 4-5 inverses. These are fun and easy :)

INVERSES:

****FIRST THING YOU MUST DO IS GRAPH!
*use the horizontal line test to see if it is an inverse. If it only goes through the line once it is an inverse. if it goes through the line more than once it isnt an inverse.
**switch x and y.
*solve for y.

I left my notes at home so im doing this off the top of my head. I hope this stuff is right. And leaving my notebook at home does not help the fact that we have the other part of the chapter test tomorrow :(
I really do not want to go to school tomorrow. Ive been doing schoolwork allllllllll day and i dont think i will ever finish. It makes me really sad that this is what school has done to me. Free time? what is that? All it does is stress me out!

4-5

We learned section 4-5 to complete the rest of chapter 4 before we go back to trig. It was all about finding inverses of a problem. It was very simple and easy because all you have to do is switch the x’s and y’s in a problem and then solve for y. There are very few steps and it is pretty simple. BEFORE you solve the problem, you need to graph the equation on a coordinate plane and use the horizontal line test.

The horizontal line test is where you draw horizontal lines across a graph. If it does NOT touch more than one point, then it is an inverse. According to the book, if it does not pass this test, then an inverse does not exist. We did learn though that in college, we will be told something different. Also, when you find the inverse, you put an exponent of -1 after the y to show that it is an inverse.

Once you passed the horizontal line test, you can get to working the problem:

Example:

Find the inverse for y = x -9
You graph this equation, and do the horizontal line test, and it does pass.
1. Switch the x and y. x= y-9
2. Solve for y. y = x + 9
3. Y^-1 = x + 9

4-5

well to be honest i really dont remember much of this week. This is probally due to the four day weekend we had. We took a test friday, but i really dont remember anything before that. But anywho, i do remember section four five,. This was one if the harder sections of the chapter. Section four five is on inverses.The rules that you have to follow are simple..
1.Draw or sketch the graph
- x^2= parabola
-x= a line
-x^3 = this line..
-sqrt x=that thing..
- ab. value = "v"

* if the graph passes the horoizontal line test, it has an inverse.

2. next you switch the x's and the y's
3. Solve for y

ex.
f(x)=x^2+3x
1.has no inverse because it does not pass the horizontal line test.

ex. y=5x+3
1. passes the horizontal line test
2.x=5y+3
x-3=5y
y(-1)=1/5x-3/5 <---inverse

ex. y=l3xl -6
1. does not pass horizontal line test

4-5

Inverses

Section 4-5 was about finding inverses. To find an inverse you have to use the horizontal line test. This is similar to the vertical line test. If it passes without touching two points then there is an inverse that is a function. That is the first step - drawing the graph.

After that, to find an inverse:

1. Switch x and y.

2. Solve for y.

To check if something is an inverse:

f(g(x)) = x and g(f(x)) = x

EXAMPLE 1: y = x + 4

A). First, graph the equation.

B). By using the horizontal line test, we know that there is an inverse.

C). Now, switch the x and y.

D). x = y + 4 and solve for y.

E). y^-1 = x - 4

* Place -1 above y to show that it is the inverse.

EXAMPLE 2: Prove that f(x) = x - 2 and f^-1(x) = x + 2 are inverses.

A). f(f^-1(x)) = (x + 2) - 2

= x

B). f^-1(f(x)) = ( x - 2) + 2

= x

C). This proves that these are inverses.

Week 3 Blog Prompt

When you are finding the domain and range of a problem what is it telling you about the graph?

Chapter 4

This week in advanced math we were learning chapter 4 which isn't that hard. But you do need to know the steps and the different types of problems to use in order to succeed. You also need to know things like the line test to tell if a graph is a function or not. And you need to know how to find zeroes.

Horizontal line test

* The horizontal line test is used to tell if a graph is a function.

If you make a straight line going vertical and the vertical line touches the graphed line twice then it is not a funcion. In other words if X is repeated then it is not a function.

Ex: (3,3) (4,3) (5,3) and (6,3) are the points then this is a function because X doesn't repeat itself.

On the other hand (3,3) (4,3) (5,3) and (3,6) would not be a function because x repeats itself therefore it would not pass the vertical line test.

blogg..

Well, I have not been feeling too well this weekend and i almost forgot about blogging! so that is why i am so late, but anyways.. This week Mrs. Robinson was out a lot , I think it was two days. While she was out we did worksheets on domain and range; which is not so bad, and on factoring which is pretty simple. Ill blog on Domain and Range because many people seemed to have problems with this.
Ex.
x^5+3x-2

first, we figure out whether the equation is a polynomial, an absolute value, a fraction, or a square root. Because the equation does not have a variable in the denominator it therefore is a polynomial.

the domain of a polynomial is always (-infinity, infinity) <-- you have to pay attention to the parenthesis also. Parenthesis only go with infinities and fractions.

The Domain of example one is (-infinity, infinity)
the range is also (-infinity, infinity)

when the exponent of the polynomial ( the greater one) is even the range is (-infinity, infinity) it it is even, then it would be the [number, then infinity)


ex.2 l-4xl +3

the domain of an absolute value is (-infinity, infinity)
the range is the [vertical shift, infinity)

4-3

This past week in advanced math Mrs. Robinson was out for two days so we were given some worksheets to work on. Looking back to the week before last we learned some more of chapter four. Let’s review section 4-3 which deals with symmetry. There are a few rule you need to know:
Check to see if:
X - axis – 1) Plug in (-y)
2) Simplify and solve for y
3) If equations are equal then it has symmetry about x.

Y – axis – 1) Plug in (-x)
2) “ “
3) “ “ y – axis

Origin – 1) Plug in (-x) and (-y)
2) “ “
3) “ “ origin

Y = X – 1) Switch x and y
2) Solve for y
3) “ “ y = x

To graph these types of equations you will do 1 of 3 things:

A) Sketch – f(x) Reflect the x-axis
B) Sketch – f(-x) Reflect the y-axis
C) Sketch - |f(x)| Flip above the axis

Tired

Yeah I'm not in the mood to do a blog, but my parents say if they keep seeing 0's on edline, well i can't tell you because that's how horrible it will be. Anyway, back to Math. This week Mrs. B-Rob wasn't here for most of the week yet she still managed to give us a quiz :). Fortunately, this quiz was generally kind of easy except for the domain and Range part. I kind of forgot to pay attention that day. We also learned some other stuff about symmetry. This is pretty easy as well, I remembered to pay attention that day, but like all Math the problems can be made extremely difficult by adding in other stuff like extra letters and exponents, etc. Hopefully this week is a week I feel like paying attention so that I can do good on tests. Susan that sucks about your computer, Dance Team I hope ya'll did good, and everyone else, PEACE.

4-3

This week in school we learned chapter 4 section 3. In 4-3 we are learned about symmetry and how to plug it in on a graph. I can have example in this sections cause there is no way for me to make a graph and stuff. I will explain how to do the problems in section 4-3.

x axis: plug in (-y), simplify( solve for y), if the equation are = then it has symmetry a/b x.
If it gives you a graph and tells you to sketch -f(x), you take the graph and draw everything on the x axis on the opposite side.


y axis: plug in (-x),simplify (solve for x), If equations are equal then it has symmetry on the y axis.
If it tells you to sketch f(-x) then you take everything on one side of the y axis and draw it to the other.
|f(x)|, flip above axis
1. plug in (-x) & (-y)
2. simplify
If it tells you to sketch |f(x)| then everything on the lower half of the axis, you draw onto the top half.

y=x is also a inverse.

Ms.Robinson this is all i can explain lmao this section is soo confusing that is why i did it to understand it more.

Blah

So, here I am, finally remembering to do my blog. Sigh. I really don't understand why my memory is so terrible.
Anyway, this week was fairly simple in Advanced Math. This section had to do with notation. There are four formulas to find notation: sum, difference, product, and quotient of functions.
1. Sum of f and g: (f+g)(x) = f(x) + g(x)
2. Difference of f and g: (f-g)(x) = f(x) – g(x)
3. Product of f and g: (f * g) g)(x) = f(x) * g(x)
4. Quotient of f and g: (f/g)(x) = f(x)/g(x), provided g(x) not equaled to 0
All this is basically saying is that in the problems that follow, plug everything in the parenthesis of f or g into the equation where there is a variable

For example,
f(3)= 7x= 21
You'd then plug 3 in where there's a x.
7 * 3 = 21
Ta da

For another example
g(4)= 5-x= 9
Then you plug in 4 for the x.
5 - 4 = 9

4-3

sooo my computer has a virus and wont let mee open up anything other than the internet, so I have no idea how many words this is going to be.

Pretty much everything except for 4-2 that we have learned this week has to do with drawing graphs, and since I already did a blog on that and I obviously can't draw any graphs on here, I'll just explain how to work some problems from section 4-3. This section was all about symmetry. Some problems gave you a graph and you had to either sketch:
-f(x), reflect the x axis
1. plug in (-y)
2. simplify (solve for y)
3. if eqations are equal then it has symmetry about x.
* If it gives you a graph and tells you to sketch -f(x), you take the graph and draw everything on the x axis on the opposite side.
f(-x), reflect the y axis
1. plug in (-x)
2. simplify (solve for x)
3. If equations are equal then it has ymmetry on the y axis.
* If it tells you to sketch f(-x) then you take everything on one side of the y axis and draw it to the other.
|f(x)|, flip above axis
1. plug in (-x) & (-y)
2. simplify
* If it tells you to sketch |f(x)| then everything on the lower half of the axis, you draw onto the top half.

Sorry this was confusing, but it is hard to explain it without having some graphs as examples.

So i'd rather not spend my friday night blogging but i wont be home all weekend so here i go.....

4-2 was notation:

1. sum of f and g=(f+g)(x)=f(x)+g(x)
2. difference of f and g=(f-g)(x)=f(x)-g(x)
3. product of f and g=(f x g)(x)=f(x)-g(x)
4. quotient of f and g=(f/g)(x)=f(x)/g(x)

most fuctions of notation are composition functions: which is a function inside another function.

4-3 symmetry:

*you get to draw graphs!

x-axis:
1. plug in (-y)
2. simplify=solve for y
3. if equations are equal then it has symmetry about x.

y-axis:
1. plug in (-x)
2. simplify=solve for y
3. if equations are equal then it has symmetry about y.

origin:
1. plug in (-x) and (-y)
2. simplify=solve for y
3. if equations are equal then it has symmetry of the origin.

y=x
1. switch x and y
2. solve for y
3. if equations are equal then it has symmetry for y=x

*when you draw these graphs its all about reflecting.

EVERYONE ENJOY YOUR 3 DAY WEEKEND :)

4-2

I’m posting this super early, but it’s because I won’t be home till late Sunday night! Anyways, Section 4-2 dealt with notation. There are four formulas to find notation: sum, difference, product, and quotient of functions.
1. Sum of f and g: (f+g)(x) = f(x) + g(x)
2. Difference of f and g: (f-g)(x) = f(x) – g(x)
3. Product of f and g: (f x g)(x) = f(x) x g(x)
4. Quotient of f and g: (f/g)(x) = f(x)/g(x), provided g(x) not equaled to 0
*f(x) is a notation!

EXAMPLE 1: f(x) = 2x + 1 and g(x) = 2 – x Find the sum and difference notations.
A). (f+g)(x) = (2x+1) + (2-x)
= x + 3
B). (f-g)(x) = (2x+1) – (2-x)
= 3x -1

EXAMPLE 2: If you have f(#), f(y), or f(i^2), the notations means to plug what is in the parenthesis into the equation instead of x. Knowing this, solve (f+g)(3) using Example 1’s f(x) and g(x).
A). (f+g)(3) = (2x+1) + (2-x)
= x + 3
B). Plug in 3 into the above answer.
= 3 + 3
C). Your answer is 6.

4-1

This week in advanced math we started chapter 4, we dealt with section 4-1 which was Domain and Range. We also had the vertical line test. But 4-1 involves 3 different types of problems. Polynomials, Square Roots, and Fractions. The secret behind chapter 4 is that you have to pay attention to see which formula you have to use.

4-1 Polynomial

-Domain is always (-∞, ∞)

-Range os (-∞, ∞) if odd

Ex: f(x)= 25-5x; solve for x and find domain and range

0= 25-5x

5x=25

x=5

D: (-∞, ∞)

R: (-∞, ∞)