REFLECTION #30

For this week I'm just going to review a few things from Chapter 6 since we're probably taking that test on Monday. Maybe this will help me remember how to do all this stuff..ha So here are some examples of concepts in chapter 6.

Ex. 1.) Give the equation for each circle:
A.) C(4,3) r=2
*Okay so you're given the center> (4,3) and the radius, 2.
*And all you have to do is plug the numbers into this formula:
(x-h)^2 + (y-k)^2 = r^2 ..(where the center is (h,k))
*So you get (x-4)^2 + (y-3)^2 = 4
B.) C(-4,9) r=3
*againg you're given the center and the radius, but this time the "h" of the center is negative, so the sign in the equation will be different
*So you get (x+4)^2 + (y-9)^2 = 9
**And remember, if you are given a coordinate for the center with zero in it, you will leave it as "x^2" or "y^2" instead of with the parenthesis.

Ex. 2.) Give the equation of the circle..
x^2 + y^2 - 2x - 8y + 16 = 0
*Since this is an equation, you're first going to have to separate the x's and y's. Then you're going to do something sorta like completing the square where you leave a space and add numbers in.
*So you first set it up like this:
x^2 - 2x + ____ + y^2 - 8y + ___ = -16
*Then you do like completing the square and take the "b" terms divide them by 2 and square them
*so you get (-2/2)^2 = 1 and (-8/2)^2 = 16
Then you add those numbers into the blank spaces of the equation *and to 16 on the other side of the equal sign* then simplify it
*So you get:
(x-1)^2 + (y-4)^2 = 1 <
Ex. 3.) Here's an equation of an ellipse. Find the following things listed..
x^2/4 + y^2/9 = 1

A.) center - (0,0)
B.) major axis - y axis, because it has the bigger denominator
C.) minor axis - x axis, because it has the smaller denominator
D.) Vertex - to find this you take +/- sqrt of the big denominator. and that would be the y axis soooo your coordinates of the vertex are (0,3)(0,-3) *because you take the sqrt of 9
E.) other intercepts - *take the sqrt of the other denominator, 4. And its the x axis sooo your coordinates are (2,0)(-2,0)
F.) focus - *for this you set up this equation and solve: smaller denom = larger denom minus focus^2
*So you get 4 = 9 - c^2
subtract 9 over
-5 = -c^2
divide by -1
And you get that c is +/- sqrt of 5 *and focus deals with the major axis, so it's in the y axis
*so your focus is (0,sqrt of 5)(0,-sqrt of 5)
G.) length of major axis - *for this you use 2(sqrt of big denom)
So you get 2sqrt of 9, which equals 2(3) and that gives you 6.
H.) length of minor axis - *use 2(sqrt of small denom)
So you get 2(sqrt of 4), which equals 2(2) and that equals 4.
*and then you would graph it, but I'm not doing that on here soo..

Okay I'm done reviewing for the week (:

im going to review trigonometry. we learned a bunch of stuff about trig functions, and I'm gunna help you out with trig functions. Cos, Tan, Csc they're all generally really easy. Study all ways to find trig functions. Period. Then study your trig chart.. A lot. Tonight I'm going to explain how to find all of the trig functions by being told only a point on a graph.The first thing you need to know if your trig functions and what they are equal to:sin*= y/rcos*= x/rtan*= y/xcsc*= r/ysec*= r/xcot*= x/y*denotes thetaok now you are given a point...let's say, (3,4)Now you would usually draw a graph but i can't do that on a computer so here's the shortcut:Let 3=a and 4=b. On the graph you make a right triangle. So now that you have a and b what are you going to do? That's right, use your formula for to find the hypotenuse of a right triangle: a^2 + b^2 = c^2So your hypotenuse = 5now, x=3 y=4 and r=5 because your hypotenuse is r.Now you plug that into the handy dandy trig functions you memorized. So:sin=4/5cos=3/4tan=4/3csc=5/4sec=4/3cot=3/4Thats all there really is to it. Its really simple. Get it? something that i don't understand is how to figure out triangles.

reflection

LAW OF COSINES

(opp leg)^2 = (adj leg)^2 + (other adj leg)^2 - 2(adj leg)(adj leg) cos (angle b/w)

You can use an angle to orient yourself like SOHCAHTOA


Example:

a is 5 cm
b is 6 cm
angle C is 36 degrees
find c

c^2 = 5^2 + 6^2 - 2(5)(6) cos(36 degrees)
c = sqrt of ((6^2 + 5^2 - 2(5)(6) cos 36 degrees))
c = 3.53



TRIG INVERSES

you have to find 2 angles usually with trig inverses.
first you need to determine which quadrants your angles are going to be in
(find out where that function is in its sign--ex: sine would be positive in the upper two quadrants, I and II)
then you plug in the positive inverse value into your calculator to get a reference angle, if there is not one given already

quadrant rules:
I to II---make it negative and add 180 degrees
I to III---add 180 degrees
I to IV---make it negative and add 360 degrees
II to IV---add 180 degrees

Example:

cos^-1 (-1/2)

Cosine is negative in the II and III quadrants
Your reference angle is given as 60 degrees.
I to II---make 60 negative and add 180 degrees = 120 degrees
I to III---add 180 degrees to 60 degrees = 240 degrees

Reflection #29?

Alright, so this week was another crazy with with no school on monday and a field trip thursday. Now I am completely lost in this class..ha. One thing I kind of understand now is 11-1.
Soo, my weekend was filled with nausea, heartburn, and indigestion. Where's the Pepto Bismol?
I'll try to explain a little of 11-1.


I'll explain how to give the polar coordinates of something.
When do this you have to take the numbers it gives you and plug it into the formula r=+/- Square root of x^2+y^2.
Then plug in the coordinates again into theta=inverse of tan (y/x)
Type it into your calc. and then figure out what quadrant the number is in, and which ones you have to move to. Get your two numbers then put them into coordinates with the number you found by doing r=+/- Square root of x^2+y^2 (being your x)

Example:

Give the polar coordinates for (3,4)

r=+/- Square root of 3^2+4^2= +/- 5
Theta=tan inverse (4/3)
Quadrants 1 & 3: 53.130 & 233.130
Answer= (5, 53.130) (-5,233.130)

*This is converting rectangular to polar.

reflection

Trig Inverses!!! ahhhh!!!

first you need to determine which quadrants your angles are going to be in
then you plug in the positive inverse value into your calculator to get a reference angle, if there is not one given already

quadrant rules:
I to II---make it negative and add 180 degrees
I to III---add 180 degrees
I to IV---make it negative and add 360 degrees
II to IV---add 180 degrees

Example:

cos^-1 (-1/2)

Cosine is negative in the II and III quadrants
Your reference angle is given as 60 degrees.
I to II---make 60 negative and add 180 degrees = 120 degrees
I to III---add 180 degrees to 60 degrees = 240 degrees

i think thats pretty simple:)

Reflection 3/7

This is some review when we started doing trig problems.

For Area say you have a triangle with:

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

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

A= 10 Sin 30 degrees which aproximately = 5.


For another triangle:

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

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

A= 12 Sin 60 degrees which aproximately = 10.392

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

reflection

flatland is gay

SOHCAHTOA:
sin (angle)=Opposite leg/Hypotenuse
cos (angle)=Adjacent leg/Hypotenuse
tan (angle)=Opposite leg/Adjacent leg

very simple, right?
memorize this and you should be fine, or atleast for all the right triangles!!!
now for non-right triangles, I'll explain how to find their area!

A=1/2 (leg) (leg) sin(angle between)
A=1/2 (2) (3) sin(35)A is approximately equal to 1.721!

Reflection march 7

Identities: MEMORIZE them. If you dont you cannot pass the identity parts of tests. I'll give an example problem.
simplify: tanA x cosA
because we memorized our identities we know that
tan=sin/cos
so now its sinA/cosA x cosA.........which equals sinAcosA/cosA...cos cancels, your answer is sinA. Yay.

another: (1-sinA)(1+sinA)foil it out....
1 - sinA + sinA - sin^2A=1-sin^2Aonce again, our identities give us an equation,
which is sin^2 + cos^2 = 1.
solve for 1-sin^2 and you get cos^2, so guess what........
you're right, the answer to this example is cos^2A.

yet another: sinAcosA
identities let us change cos and we get:
sinA cosA/sinA, simplify, and cosA is your answer.

Don't forget to take your time with these problems and remeber, do identities, then algebra, then identities, then algebra, and keep continueing this until it is fully simplified.

Reflection, March 7th

I'm gonna review something we learned at the very begining of trig, SOHCAHTOA.

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

Here are some examples on using SOHCAHTOA....

solve for b & c.

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

which equals about 75.229

sin28=40/c
c=40/sin28

which equals about 85.202

REFLECTION!!!

for trig inverses find 2 angles

first you need to determine which quadrants your angles are going to be in
then you plug in the positive inverse value into your calculator to get a reference angle, if there is not one given already

quadrant rules:
I to II---make it negative and add 180 degrees
I to III---add 180 degrees
I to IV---make it negative and add 360 degrees
II to IV---add 180 degrees

Example:

cos^-1 (-1/2)

Cosine is negative in the II and III quadrants
Your reference angle is given as 60 degrees.
I to II---make 60 negative and add 180 degrees = 120 degrees
I to III---add 180 degrees to 60 degrees = 240 degrees

REFLECTION #29

Okay I think I'll just review some things that were on the Chapter 2 test we took recently. Soooooo here's a few examples.

Ex. 1.) Factor: 6x^2 + 18x + 3x + 9
*Since this problem has an even amount of terms you can see if you can factor it by grouping.
*so you group it like this> (6x^2+18x)(3x+9)
*Now look at the first parenthesis and see what you can factor out of it.
*you can factor out a 6x, and you're left with 6x(x+3)
*Now look at the other side and see what you can factor out. you can factor out a 3.
*then you get 3(x+3)
*Soo all together you have 6x(x+3)+3(x+3)
*Then you get (x+3)(6x+3)
*Now all you do is set what's inside each parenthesis equal to zero and solve
*So you get x+3=0. subtract 3 over. so you get -3
*Then 6x+3=0. subtract 3 and divide by 6. So you get -1/2
*So your final answers in point form are (-3,0)(-1/2,0)

Ex. 2.) Factor: 6x^2 + 11x - 35
*First decide which method of factoring you should use
*The problem is not factorable so you have to either use completing the square or the quadratic formula
*Since the "b" term is odd, you should use the quadratic formula
*So you use this formula: x=-b+/sqrt of b^2-4ac/2a
*So you get x = -11 +/- sqrt of (11)^2 - 4(6)(-35)/2(6)
*x = -11 +/- sqrt of 121 + 840/12
*x = -11 +/- sqrt of 961/12
**Then you get that x = -11 +/- 31 / 12
*When you don't have a squareroot in the answer, that means you have to solve the rest of this problem in two ways.
*First you solve -11 + 31/12 to get your first answer. you get -42/12 which reduces to -7/2
*Then to get your second answer you take -11 - 31/12 and that gives you 20/12 which reduces to 5/3
*So your final answers in point form are (-7/2,0)(5/3,0)

Ex. 3.) x^2 + 6x + 10 = 0
*This equation is not factorable so again you will have to use either the quadratic formula of completing the square.
*In this case, the "b" term is even so you should use completing the square
*So first you subtract 10 over to the other side
*Then leave a space between 6x and the equal sign
*So you get x^2 + 6x = -10
*Now take your "b" term, which is 6, divide it by 2 and square it.
*So you get 9. Now add 9 to the empty space and add 9 to -10
*So you get x^2 + 6x + 9 = -10 + 9
*Now simplifying that you get (x+3)^2 = -1
*Now take the square root of both sides.
*So you get x + 3 = +/- i
*Then subtract the 3 over.
*So you get x = -3 +/- i
*So in point form your answer would be (-3 + i, 0)(-3 - i, 0)

Alright I think that's enough..

Reflection #29

Okay, I'm going to review word problems. We had one on our ch. 2 test we took Thursday and it's been so long since we did them...

A word problem:
"I am building a rectangular pen to keep all my pigs from escaping the farm. I sue the side of my outhouse as part of the enclosure and only have 100ft of wood to use. What is the mas. area I can enclose for my pigs?"

1. draw figure
2. the width of either side is x and the length is 100-x/2
*A=length times width
3. 100=2x x=50
4. 100-(50)/2 =25
5. 25 x 50
6. 1250ft^2

This isn't the one we had on the test, but it's the same concept. If you have any other examples, feel free to post!

Reflection 3/7

So i had a three day week this week, because of teacher stuff on monday, and softball on friday, so...i think i missed a chapter test on friday, so i guess i will have to make it up on monday. So onto Flatland, we watched the movie on Tuesday, hmmmmm. That is all i can say. The book was sort of different from the movie, because in the movie A. Square had a granddaughter and in the book he had a grandson. So there are some differences in the movie/book. So we started chapter tests on wednesday, and i am surprised on how much i know on the tests. I thought i would not know a thing, but I do not think that i did that terrible. I don't know how to do my blogs anymore, because there is nothing to blog about, except that we are taking chapter tests throughout the week. And i also don't know how to comment on people's blogs because they are blogging about the same thing as i am, nothing. hahah. Hopefully this week is a good week and i bring all of my grades up before report cards. kbye :)

reflection march 7

So we finished flatland last week and we watched the movie tuesday. The movie and the book are not even the same, the movie had a grandaughter and the book had a gandson. So yeah, they had their differences, but i pretty much got the visual picture of what the book was talking about. Wednesday we started doing our practice tests which i didn't really know anything! ha. And then i missed friday so i'll have to do 2 tests tomorrow, ugh. I guess i have to look over my notes for chapters 3 and 4. I thought that these tests we going to be easy, but i don't really remember much. I wish we could use our notebooks or work in groups:/ oh well. I'll just review my notes and hopefully i'll remember everything for tomorrow. And another thing..i don't even know what to comment on during the week anymore. Most people's blogs are just kind of like mine so i don't know how to comment on that. So anyway, back to school tomorrow..ugh, we actually have a full week of school this week.

3/7/10

so another four day week, and all we really did was review chapters and watch flatland. the movie itself seemed really wierd and was like a drug trip, but it helped me understand the book more. we also repeated the chapter 1 and 2 tests, which jogged my memory.
the movie followed up on the book really close. the people in flatland used the 3 ways to determine who is there, the concept of 2 dimensions was easy to understand, and the part where the sphere showed up started to get a little hard to follow because of all the movement going on. the main difference i saw was that the women were not lines, and A. Squares grandchild was female also
i'm sure tomorrow we will have another chapter test, so i should go look over chapter 3 and 4

reflection for 3/7

ive lost count of the blogs nowadays, everybody has different numbers now, so theres no actual way to find out what number it is

anyways, we didnt really do much this week either, but it WAS a 4 day week and we did watch flatland on tuesday and did nothing on friday, so......idk, i guess ill just put some math that i can remember

here's some sequences and series(es), which are pretty easy if u kno what ur doin:

Find the formula for the nth term of the arithmetic sequence. 3,5,7...
tn=3+(n-1)(2)
tn=3+2n-2
tn=1+2n

Find the formula for the nth term of the sequence. 3,4.5,6.75...
r=4.5/3=3/2
6.75/4.5=3/2

tn=3x(3/2)^n-1


thats all ive got for now.....bein sick sucks.....blah