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.