Reflection #35

A review of something we did this week: DETERMINANTS

-to find the determinant of a 3x3 matrix, delete a row and a column while alternating signs
*pick a row and column with the most 0's and 1's

Example:
[1 2 3]
[4 5 6]
[0 0 1]
**that's suppose to be one whole matrix
-the first row/column you should choose is the corner 0, then the second 0, then the 1
=0 [2 3] - 0 [1 3] + 1[1 2]
.....[5 6]......[4 6]......[4 5]
-when you multiply by the 0's, you get 0, so you're left with the 1
-to find the determinant of the last matrix, just cross multiply 1 by 5 and 2 by 4, subtract the two products, and multiply them by 1
=1[1 2] = 1(5-8) = 1(-3) = -3
....[4 5]
-your answer is -3

Reflection

This week we did review of trig but did also do stuff with chapter 12. While doing my review packets i decided to give examples on here from chapter 7. I found the easiest thing with chapter 7 was the conversion from degrees to radians and radians to degrees. Also, converting degrees to degrees, minutes, seconds was simple. The trig chart i have basically memorized by now, but overall pretty simple to comprehend. I'll give the unit circle then a few conversions for examples.

EXAMPLES:

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

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

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

So if anyone wants to help me with the chapter 9 review packet that would be great b/c its the one i'm having the most trouble with. Thanks :)

late blog

im going to review Completing the Square when completng the square the first thing you have to di is put all X's to one side of the equation and your whole numbers to the other side of the equation. then you make your quadratic term equal to 1(it should be x^2). then you take your binomial co eff. and pit it in this formula (-b/2(a)^2. you multiply your quadratic co eff. by two and then you divide. after you come up with the first answer you square it. then you plug the answer into the problem on both sides of the equation x^2+x+1=3+1. then you simplify. ageter that you take the answer you had before and you square it and put it in another equation and it would be (x+1)^2=4. the you square root bothe sides and solve for x and your final answer would be x=3.

Homework

Quick question: What was all the homework assignments for ch. 12? (pages and problem numbers)

please and thank you!

Easter Sunday Blog

The steps up to graphing an ellipse:

Formula:
(x-h)^2/(length of x/2)^2+(y-k)^2/(length of y/2)^2=1

1. center: (h,k)

2. major axis: larger denominator

3. vertex: on major axis; +/- square root of small denominator

4. find focus: smaller # or denominator^2=larger # or denominator^2-focus^2

5. length of major axis 2 square root of larger denominator

6. length of minor axis 2 square root of smaller denominator

7.then graph...

Reflection 4/11

The Area of a right triangle and non-right triangle:

Area of a right triangle: A=(1/2)(base)(height)
Area of a non-right triangle: A=(1/2)(leg)(other leg)sin(angle in between the two legs)

Examples:
visualize a right triangle: DEF (left to right)
D=90 degrees, d=8, e=6...Find the area.
So, you know your base is e, which is 6.
Now, to find your height, or f, use: a^2 + b^2 = c^2
c is always to hypotenuse, which is d (8).
6^2 + b^2 = 8^2
36 + b^2 = 64
b^2 = 28
b = 5.292
Your height is 5.292
Now, plug into the formula.
A = (1/2)(6)(5.292)
A = 15.875

Spring Break also sucked, beside not being at school

Reflection #34

When we were doing those packets, one of the things I had trouble with was identities, so that's what I'm going to review about this week:

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

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

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

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

I think I've got all the identities, but if I forgot any, please comment...

Reflection 4/11

Spring break didn't last long enough, :(

Here are some formulas that we learned in chapter 10:


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

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

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

This is also some other Laws that we learned in that chapter:

LAW OF SINES

sinA/a = sinB/b = sinC/c

LAW OF COSINES

(opp leg)^2 = (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg)cos(angle between)



Example:

x= |6^2 + 5^2 -2(5)(6) cos 36*

x=3.530

________________________________________________________

Here is the change of base formulas:

Changing Bases
....use when a log cannot be solved
....used to solve for x as a variable
....used to change the base of log

1. exponetial form
2. take log of both sides
3. move exponents to the front
4. solve for variable
5. write as a function or whole number

Example:

log base 3 of 7
1. 3^x=7
2. log 3^x = log 7
3. x log 3= log 7
________________________________________________________

We also learned what the different quadrants mean, and how to switch from each of them:

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

Why do we have to go back to school. :/

law of cosines
(opp leg)=(adj leg)^2 + (other adj leg)-2(adj leg) (adj leg) cos (angle between)

it is used when solving for non-right triangles
SOHCAHTOA is also the law of sines
law of sines
EX: angle B=30 degrees, angle A=135 degrees, and side b=4 ...find C and a and c.
find your other angle, so add 30 to 135 together and get 165, then you subtract that from 180 and get 15 so angle C is 15 degrees.
next, find another side...since you have pairs, you ca use that in the law of sines
your equation would be sin(30/4)=sin(135/a)
cross multiple and get asin(30)= 4sin15........which means that c=2.071

Reflection

The Unit Circle

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

Trig Chart:

0°
sin0=0
cos0=1
tan0=undefined
sec0=1
cot0=0

30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3

45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1

60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2

90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefinedcscπ/2=1
secπ/2=undefined
cotπ/2=0

Reflection 4/11

okayyy, so we have school tomorrow :/ and i know that its a bummer! But at least we have B-rob back and we can actually learn math. I actually understand what we are learning this time around. But when we get back to school, i'm going to have to review, because i might have forgot some stufff.

________________________________________________________

EXAMPLES:

(a) velocity = distance/time speed= /velocity/

velocity = (-9,12)/3 = (-3,4)

speed= squareroot -3^2 + 4^2 = squareroot 25 = 5

(b) (x,y) = (5,3) + t(-3,4)

(c) How far does the object travel after 5 seconds.

(x,y) = (5,3) + 5(-3,4)
(5,3) + (-15,20)

= (-10,23)


(d) Write in parametic equations...

x=5t-3t
y=3+4t
___________________________________________________

**REMEMBER YOUR IDENTITIES:

csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
_______________________________________________________

IF anyone wants to show me some more examples to review, i loveeee youu! haha, THANKSSS :)

Reflection

im going to reveiw t domain, range, functions, finding any symmetry, and inverses. What I found was the easiest to learn was finding the symmetry.EXAMPLE: y=x^3=4xa.) x-axis(-y)=x^3+4xy=-x^3-4x Reflect; Nob.) y-axisy=(-x^3)+4(-x)y=-x^3-4x Noc.) y=xx=y^3+4yy^3+4y=xy(y^2+4)=x Nod.) origin(-y)=(-x)^3+4(-x)-y=-x^3-4xy=x^3+4x YesI also understood how to do domain and range very well. The types of the domain and range problems i understood the most were the fractions and polynomials. (oo => stands for infiniti EXAMPLE: y=x^3+4x^2+12For any type of polynomial the domain would be (-oo,oo) and for the range, odd:(-oo,oo); quadratics: [vertex,oo] or [-oo,vertex].The answer to this problem would be Domain: (-oo,oo) Range (-oo,oo)EXAMPLE: 5x+4/x^2-4First, set the bottom of the fraction equal to zero.x^2-4=0+4+4x^2=4x=+ or - 2Your answer then comes to,Domain: (-oo,-2)u(-2,2)u(2,oo)What I am having the most trouble understanding is the inverses. I get most of the concept, but I'm still getting confused. Can anyone help me with that??

Reflection

this if from when we learned how to use exponents, logs, and graph equations. the thing i understood the most was the logs. here's how you would work one:log base b x=astep 1:when putting this in exponential form you take the base set it rased to what the problem equals and set this equation equal to x.b^a=xstep 2: this is exponential form the problem can no further be simplifyed.***if given the equation log base 2 16, remember that if a equation is not set equal to anything set it equal to x.******if given the quation log 1020=x, remember that when there is not a base it is 10.***heres an example:log base 2 16=x2^x=162^x=2^4x=4log base 2 16=4heres some thing i didn't understand (b^2/a)^-2(a^2/b)^-3 i don't understand the sandwitch thing and i don't realy get how solve this problem. some help would be nice thanks..

Reflection

this if from when we learned how to use exponents, logs, and graph equations. the thing i understood the most was the logs. here's how you would work one:log base b x=astep 1:when putting this in exponential form you take the base set it rased to what the problem equals and set this equation equal to x.b^a=xstep 2: this is exponential form the problem can no further be simplifyed.***if given the equation log base 2 16, remember that if a equation is not set equal to anything set it equal to x.******if given the quation log 1020=x, remember that when there is not a base it is 10.***heres an example:log base 2 16=x2^x=162^x=2^4x=4log base 2 16=4heres some thing i didn't understand (b^2/a)^-2(a^2/b)^-3 i don't understand the sandwitch thing and i don't realy get how solve this problem. some help would be nice thanks..

reflection 4/11

this week was pretty fun haha since it was spring break and all... :/ haha time for school.

sequences and series:

arithmetic series: 3,5,7,9...
geometric series: 2,4,8,16...

arithmetic formula: a.n = a.1 + (n-1)d
geometric formula: a.n = a.1 x r^(n-1)

you can solve for any term number by plugging in the numbers to the formula, its not that difficult and i liked this chapter

reflection 4/11

so spring break is finally over. it went by pretty fast, and its time to focus on school again...

11-2

Z = a + bi Regular form

z = rcosO + rsinOi Polar
aka z = rcisO

|z| = sqrt. a^2 + b^2

To multiply complex numbers as rectangular - FOIL
To multiply as polar - multiply rs, add the angles

4cis45*
4cos45* + 4sin45*i

12cis5*
12cos5* + 12sin5*i

Reflection

So we have to go back to school tomorrow..that sucksss! I'm gonna go over law of sines, which is somewhere around chapter 8 or 9, It'll probably be on our trig exam. This formula may be useful in doing this: sinA/a=sinB/b=sinC/c. This formula is used when you know pairs in non right triangles. You are setting up a proportion, to make it easier. Heres some examples:

1) An ABC triangle with b=123, a=16, and angleB= 115 degrees. Find A.

sin115/123=sinA/16
16sin115=123sinA
sinA=16sin115/123

A=sin-1((16sin115)/(123))
A= about 6.771 degrees


2) An ABC triangle with angleA=30 degrees, angleC=25 degrees, and a=4. Find c.

sin30/4=sin25/c
csin30=4sin25
c=4sin25/sin30
c= about 3.381 degrees

-----------------------------------

I understood everything we learned before we left for the holidays so if someone could just give me an example of anything we learned in trig that would be good I guess?

reflection 4/11

This is my reflection for the second sunday during spring break. This week went by soooo fast! Its so depressing:/ It didn't really feel like a break because I had to go to school every dayyyy for softball. I definately do nott feel like going to actual school tomorrow..but I guess the sooner we go back then the sooner we get out for the summer:) This blog is going to be about vector equations. I think we did this wednesday.. They are used for a moving object. Speed is the /v/ and remember that this is a magnitude. The (a,b) can also be called a direction vector.

EXAMPLES:

(a) velocity = distance/time speed= /velocity/

velocity = (-9,12)/3 = (-3,4)

speed= squareroot -3^2 + 4^2 = squareroot 25 = 5

(b) (x,y) = (5,3) + t(-3,4)

(c) How far does the object travel after 5 seconds.

(x,y) = (5,3) + 5(-3,4)
(5,3) + (-15,20)

= (-10,23)


(d) Write in parametic equations...

x=5t-3t
y=3+4t

reflection 4,11

reflection 4/4

This is my reflection for the first sunday of spring break..i completely forgot about blogs that sunday, i wasn't even thinking about school untill today. Soo, the week before spring break...i don't even remember what we did..I think it was vectors in chapter 12. First of all, a vector represents movement and to add picture vectors you attach to the tail of another vector. I can't really give examples on what this looks like, so look in your notes.

EXAMPLES:

* If u=(1,-3) and v=(2,5) find... (a) u+v , (b) u-v , (c) /2u+3v/.

(a)= (3,2)
(b)= (-1,-8)
(c)= /(2,-6) + (6,15)/
= /(8,9)/
= squareroot 8^2 + 9^2
= squareroot 145


* If A(2,4) and B(1,5) find the componeent form of AB.
1-2=-1
5-4=1

AB= (-1,1)

reflection 4/11

I'm having a great Easter break, but I'm sad that we gotta go back to school tomorrow :/ Hope everybody else had a great week off, but anyways, to math. I will explain how to find symmetry.

To find symmetry on the y-axis:
you have to plug in (-x) to every x there is

Example:
y = x^2 - 7
y = (-x)^2 - 7
y = x^2 - 7
Therefore, yes, there is symmetry.

To find symmetry on the x-axis:
you have to plug in (-y) to every y there is

Example:
y = x - 2
(-y) = x - 2
y = -x + 2
Therefore, no, there is no symmetry.

To find symmetry on the origin:
you have to plug in (-y) and (-x)

Example:
y = x^2 + 4
(-y) = (-x)^2 + 4
y = -x^2 + 4
Therefore, no, there is no symmetry.

To find symmetry on y=x:
find the inverse

Example:
y = x^2 + 4
x = y^2 + 4
y^2 = x - 4
y = sqrt(x-4)
Therefore, no, there is no symmetry.

Reflection

I'm so glad i dont have to go to school friday and had the whole week off for spring break. Here is some review stuff from chapter 8. We learned how to find the angle of inclination, which i found was really easy compared to some stuff we learn. We also learned about amplitudes, periods, vertical shifts, etc. There are some formulas we had to learn to be able to work these problems:

1.) For a line
m=tan alpha where m=slope and alpha=angle of inclination
2.) For a conic
tan 2 alpha=B/A-C
3.) For a conic if A=C then
a=pi/4

EXAMPLE:

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

Reflection 4/11

Here's some older stuff about SIGMA NOTATION....yay.

100 - 1 & 100 are called the limits of summation.
sigma n^2 - n^2 is the summand
n=1 - the bottom variable is called the index, which is K.
- To evaluate
-plug in the numbers between your limits of summation into the summand.
Adding each term to form a series.

- Examples:
Give each series in expanded form
1) 4
sigma 5K = 5+10+15+20
K=1
2) 6sigma 9+16+25+36
n=3

Express the series in sigma notation
1) 1+2+4+8+16+32
5
sigma 1X(2)^K
K=0

2) 48+24+12+6+...
infiniti
sigma 48(1/2)^c
c=0

reflection for 4/11

this week :/
didn't have school, so we didn't learn anything new in class, so I'll just go over something from the past

I guess I'll put up the unit circle and some identities and etc. I really can't think of that much math lately :/

Six trig functions:
sin: y/r
cos: x/r
tan: y/x
cot: x/y
csc: r/y
sec: r/x

Unit Circle:
90: (π/2): (0,1)
180: (π): (-1,0)
270: (3π/2): (0,-1)
360: (2π): (1,0)

And here are a couple identities:
csc x = 1/sin x
sec x = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x

thats about it, hope everybody else had a good spring break/easter holidays
I hope I'm not the only one that had a bad spring break/easter holidays

REFLECTION 4/11

Alright since we didn't have school this week, I'm just going to go over something we learned a while back. Here's a few examples of things we learned in Chapter 13:

Ex. 1.) t1=1 tn=3t(n-1)-1 (*anything in parenthesis is a subscript)
*Find the next 3 terms in the sequence
*Since you're already given the first term, now you have to find the second term. So all you do is plug in 2 wherever you see an "n" in the equation they gave you.
*So you get t2=3t(2-1)-1
And that simplifies to t2=3t(1)-1
Then you look and see what t1 is and plug it into the equation (t1 is 1)
So you get 3-1 = 2 Sooo 2 is the second term
*Now to find the third term. you plug in 3 wherever you see an n in the equation and it simplifies to 3t(2)-1 and that gives you 6-1 = 5
*And to find the fourth term you do the same thing, plug in 4 for the "n" in the equation, and you get 3t(3)-1 and that gives you 3(5)-1 = 14
**So the next three terms in this sequence would be 2,5,14

Ex. 2.) Find a formula for the given sequence: 6,10,14
*First you have to figure out whether it's arithmetic or geometric.
*It's arithmetic because you add 4 each time
*so you use the arithmetic formula tn = t1 + (n-1)d to make up your formula
*All you do is plug in the numbers that you know. you plug in 6 as t1 because it's the 1st term, and plug in 4 as 'd' because it's what's added each time
*So you get tn=6+(n-1)(4)
then you simplify it and get:
tn=2+4n and that's your formula

Ex. 3.) Find the sum of the first 10 terms of the arithmetic series where t1=3 and t10=39
*All you do is use the sum formula for arithmetic: sn=n(t1+tn)/2
And you plug in the numbers that you know
So you get Sn=10(3+39)/2
*And that simplifies to 210

Ex. 4.) Find the sum of the multiples of 3 between 1 and 1000
*The first thing you want to do is find the smallest and largest multiple of three between those numbers.
*The smallest multiple is 3 and the largest is 999
*You can use the arithmetic formula for this problem like this:
999=3+(n-1)(3)
999=3n
n=333
So now that you have 'n' you can plug all your numbers into the sum formula to get the answer
*So you get Sn=333(3+999)/2
And that gives you 166833

Alright I think that's enough of a review...I still don't really understand all the speed/velocity stuff we learned before we left for the holidays, so if you wouldn't mind explaining that would be great (: