Reflection

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

EXAMPLE:

C = 90 degrees
b = 7
c = 12

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

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

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

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

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

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


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

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

BLOG 4/21

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

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

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

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

Blog Topics

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

1. How to use the unit circle

2. How to solve for an angle

3.How to find a reference angle

late blog

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

Reflection 4/18

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

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

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

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

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

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



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

Blog 4/18/10

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

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

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

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

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

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

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

o to solve a determinate this is what you do

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

Reflec

Vectors.

DOTS OF VECTORS

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

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

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

Reflection 4/18

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

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


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


Also number 11?


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

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

In order to find the axis of symmetry: x=-b/2a

Vertex:
a. (-b/2a, f(-b/2a)
b. complete the square to get into vertex form
y=(x+a)^2+b ((A & B ARE #s))
(-a,-b)

Focus:
1/4p= coefficient of x^2 then add p.
*if opens up, add to y-value from vertex.
*if down, subtract y-value from vertex.
*if opens right, add to x-value from vertex.
*if opens left, subtract.

Directrix is p units behind vertex
always x= or y=
so subtract p.

EXAMPLE:
x^2+1
-V:(0,1)
-focus:
1/4p=1
p=1/4
-directrix:
y=1-1/4
y=3/4

Someone help me out with determinants??

Reflection 4/18

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

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

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

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

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

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


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

take the zero matrices out

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


Leaving space is also an option when solving:

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

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


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

Reflection

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

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

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

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

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

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

Reflection 4/18

Math, yay.

DOT PRODUCTS:

okay, here's an example of finding a dot product:
(2,3) (1,4)
-Use formula v1v2=x1(x2) + y1(y2)
-So jus multiply:
2x1=2
3x4=12
2 + 12 = 14
-So the answer is 14, see easy right?

FINDING THE MIDPOINT:
Finding the midpoint is really easy because you just plug into the formula.
But here's an example:
Find the midpoint of (2,2,2) and (2,4,6)
-Use the formula m = (x1+x2/2, y1+y2/2, z1+z2/2)
-so 2+2/2, 2+4/2, 2+6/2
Which equals (2, 3, 4)

See, easy!!!

reflection april 18

This week went by pretty fast i guess. I can't even remember anything that happened haha. But we have our first playoff game tomorrow at 5:30 at homeeeeeee:)...soo anyway, back to math, let me look in my notebook real quick because i do not remember anything that we did. ha. We did a lot of new stuff with vectors and 3-d stuff. And also determinants. I think i'll give examples on the 3-d stuff, with the 2 formulas- 1. Distance Formula and 2. Midpoint Formula.

EXAMPLES:

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

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


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


= (X-6)^2 +(Y+1)^2+(Z-5)^2=9



2) Find the equation of a plane that contains A(0,1,2) with an orthogonal vector (3,4,-2)

ax+by+cz=d

3x+4y+2z=0 d=3(0) + 4(1) + 2(-2)

4/18 reflection

this week went by super fast! haha but during this week we learned more about vectors and determinants

a vector is a quantity that is described by direction in a magnitude

magnitude = |u|



Dot Product:
v1 = (x1,y1)
v2 = (x2,y2)

v1 x v2 = x1x2 + y1y2

if = 0 then v1 and v2 are perpendicular
if v1 = kv2 then v1 and v2 are parallel

u = (5,-4)
v = (6,4)
w = (-10,-4)

test if U & V are perpendicular

(5)(6) + (-4)(4) does not = 0 therefore, no

Reflection 4/18

okay, so this week was pretty easy, and we didn't even have class on friday because of career day! But during the week, we learned more about vectors. We also learned about determinants. Which are pretty easy to understand! You just have to remember them from algebra.

_________________________________________________

EXAMPLE:


[1 7 9]
[4 5 6]
[0 0 1]

**you always want to start with the 0s and 1s. So start with those rows/columns.

=0 [7 9] - 0 [1 9] + 1[1 7]
...[5 6].....[4 6]....[4 5]

**you always go +/-/+/-...and so on.

-when you multiply by the 0's, you get 0, so you're left with the 1

**now solve the determinant...

5-28= -23

^^^and that is your final answer.
_________________________________________________


You would solve it the same way if you had letters (variables). Because on the test, she will give you a 4x4, with all letters so she knows you know how to solve determinants.

_________________________________________________

So I pretty much understood everything that we learned this week. Its pretty simple. The only thing that i may need help on is reviewing vectors again? hahaha. just a little fresh up! Examples will help, THANKS :))

reflection

im going to review vectors this week

a vector is a quantity that is described by direction in a magnitude

magnitude = |u|



Dot Product:
v1 = (x1,y1)
v2 = (x2,y2)

v1 x v2 = x1x2 + y1y2

if it = 0 then v1 & v2 are perpendicular
if v1 = kv2 then v1 & v2 are parallel

u = (5,-4)
v = (6,4)
w = (-10,-4)

test if U & V are perpendicular

(5)(6) + (-4)(4) does not = 0 no

i understand just about everything, but can someone refresh my memory on determinants?

REFLECTION 4/18

Okay so this week we learned a lot more with vectors. We learned dot product and cross product. We learned how to determine if vectors are parallel or perpendicular. And we learned how to find the angle in between two vectors, determinants, and some other things. This week I thought was really easy, and I understood everything we learned. So here's a few examples of what we learned this week:

Ex. 1.) Find the dot product: (3,-5) (7,4)
*To find dot product you use this formula > v1v2 = x1(x2) + y1(y2)
So all you have to do is multiply the x's together and multiply the y's together and then add them together after.
*So you get 3(7) + -5(4)
= 21-20
*So the dot product is 1

Ex. 2.) Find the angle between the two vectors (3,-4) and (3,4)
*To find the angle between two vectors you use this formula:
cos(theta) = u (dot) v / (magnitude of u)(magnitude of v)
*Okay so first you find the dot product of u and v
*So you get 3(3) + -4(4) = 9-16
*And that gives you -7, so that's the numerator of the fraction in the formula
*Now you have to find the magnitude of u
*So you get sqrt of 3^2 + (-4)^2
*So the magnitude of u is 5
*Now you find the magnitude of v
*you get sqrt of 3^2 + 4^2 and you get 5
*So when you put it all together and multiply the 5's you get:
cos(theta) = -7/25
*Remember, you're finding an angle so you have to take the inverse of cosine here
So you get theta=cos^-1(-7/25)
Cosine is negative in the 2nd and 3rd quadrants
you get 73.740 when you put cos^-1(7/25) in your calculator
*When you move from the 1st quadrant to the 2nd you get 106.260 degrees, and when you move from the 1st quadrant to the 3rd, you get 253.740 degrees. And those are your two angles

Ex. 3.) Find the midpoint of (3,8,-2) and (4,-1,2)
*you use the midpoint formula for this: m = (x1+x2/2, y1+y2/2, z1+z2/2)
*So all you do is add the x's together and divide by 2, and do the same for y and z)
*So you get M=(3+4/2, 8-1/2, -2+2/2)
*The midpoint would be (7/2,7/2,0)

Well I'm pretty sure I understood everything this week, so I don't have any questions

Reflection for 4/18

well, this week we learned more about vectors

the dot product:
u=(8,7) v=(1,4)

8(1) + 7(4) = 36

magnitude:
u=(2,6)

√(2^2 + 6^2) = 2√10

vectors are pretty easy to figure out, but i felt like an idiot when it took me a while to actually understand them :/

now determinants, i need some help with those, so if anybody's willing to help me out, thatd be much appreciated :D