So, i'll take this time to complain and rejoice in the SAME blog.
SHOUT OUT TO MALERIE BULOT WHO REMINDED ME ABOUT MY FINAL BLOG TODAY, a little late, YET BETTER THEN NOTHING!
Wow, ADVANCED MATH IS OVER. ON TO THE CALCCCC! I might fail miserably, but ya can't say i won't try!
:)
and for the second night in a row, i am still at edee's working on this spoof. we stayed here til 2am yesterday, and i have a feeling we're on the same roll this time.
i'm going to dieeeee, :/
here's my last blog!
To start off, there are two types of inequalities, regular inequalities and absolute value inequaties. In absolute value inequalities there are two different types determined by the signs if it is not an equal sign. They can either be "and" or "or" inequalities. "And" inequalities always have a less than symbol (<) and "or" inequalities always have a greater than symbol (>). Also, (something important to remember) in absolute value inequalities you always get two answers. I'm sure everyone remembers how to work these problems but here's a few examples:
Ex. 1.) 4x + 1 > 13
*you would just solve this like a regular equation even though it does not have an equal sign. *and for this problem you would only get one answer because it does not have an absolute value symbol.
*so you subtract 1 from 13 and you get 4x > 12
*divide by 4 and you get that x > 3
Ex. 2.) 2y - 4 = 12 (*assume there is an absolute value thing around 2y-4)
*the first thing you want to do with this problem is set up two equations. since in absolute value you know you will get two answers
*your first equation will be the original equation but without the absolute value symbol like this:
2y - 4 = 12
*and your second equation will be the same equation except you change the 12 to -12 like this:
2y - 4 = -12
*then you solve both equations for y and you get that y = -4 and 8
Ex. 3.) 3x - 4 + 5 < 27 (*assume there is an absolute value thing around only 3x - 4)
*the first thing you have to do is isolate the absolute value and to do that you have to get rid of the 5. so you subtract the 5 over from the 27 and your new equation is:
3x - 4 < 22
*the next thing you notice is that this equation has a < sign and that means it is an "and" equation so you have to set it up a certain way like this:
-22 < 3x - 4 < 22
*then you solve the equation first by adding 4 over to all sides then you get this:
-18 < 3x < 26
*then you divide by 3 on all sides
your final answer is: -6 < x < 26/3
Next comes slope! First of all, there are 3 different equations of slope. There's slope intercept form, point slope form, and standard form.
Slope intercept is y = mx + b.
Point slope form is y - y1 = m (x - x1).
standard form is Ax + By = C
Here are a few examples of problems:
Ex. 4.) Find the slope of the two points (4,1) (3,0)
*to solve this you use the formula for slope which is: m = y2 - y1/x2 - x1
*so you get...0 - 1/3 - 4 and that gives you -1/-1 which equals 1
Ex. 5.) Find the equation of the line 3x + 4y = 12 that is perpendicular to the point (3,2).
(*by the way, I'm not sure if I worded that problem right but hopefully you should know what I'm talking about right?)
*so the first thing you have to do is find the slope of the equation you are given. 3x+4y=12
*first you subtract 3x over and you get 4y = -3x + 12
*then you divide by 4. and you get that y = -3/4x + 3
*So your perpendicular slope of that equation is 4/3. (because you take the negative reciprocal of the orginal slope of the equation)
*Then to put it in an equation including the point you are given, you use the point slope formula.
*So your final answer is y - 2 = 4/3 (x - 3)
reference angles and exact answers. *Remember that reference angles can only be between 0 and 90 degrees. Here are some examples:
Ex. 6.) Find the reference angle of sin 236. (*assuming that there's a degree sign after 236)
*the first thing you have to do is figure out what quadrant 236 is in. it's in the 3rd quadrant.
*sin relates to the y axis and in the third quadrant, the y axis is negative so for the reference angle, sin has to be negative
*and to find the reference angle of 236 you have to subtract 180. and you get 56 degrees.
*so your reference angle is equal to -sin 56. and since this is not on the trig chart, to find the exact answer you would have to type it into your calculator.
Tuesday, May 11, 2010
3rd late reflection.
Notice, by this time I'm practically DEAD.
this blog thing is pretty tiring after a while,
and PLUS i'm doing last minute corrections to our mu alpha theta spoof. :(
On the bright side, functions were pretty simple to me so i should probably explain them:
First, let's start with the fact that if you find the function of x, or f(x), you will find y.
Adding or subtracting functions:
(f+g)(x)
f(x)+g(x)
(f-g)(x)
f(x)-g(x)
example:
f(x)=x+1
g(x)=x-4
(f+g)(x)= x+1+x-4
(f+g)(x)=2x-3
Multiplying functions:
(fxg)(x)
f(x) x g(x)
f(x)=2x+4
g(x)=4x-6
(2x+4)(4x-6)
F.O.I.L.= (8x^2-12x+16x-24)
=8x^2+4x-24
Finding a function of another function:
(fog)(x)=
f(g(x))
f(x)= 2x+4
g(x)= x-4
plug in the g(x) equation into the f(x) equation where ALL x's appear.
2(x-4)+4
distribute.
2x-8+4
(fog)(x)=2x-12
this blog thing is pretty tiring after a while,
and PLUS i'm doing last minute corrections to our mu alpha theta spoof. :(
On the bright side, functions were pretty simple to me so i should probably explain them:
First, let's start with the fact that if you find the function of x, or f(x), you will find y.
Adding or subtracting functions:
(f+g)(x)
f(x)+g(x)
(f-g)(x)
f(x)-g(x)
example:
f(x)=x+1
g(x)=x-4
(f+g)(x)= x+1+x-4
(f+g)(x)=2x-3
Multiplying functions:
(fxg)(x)
f(x) x g(x)
f(x)=2x+4
g(x)=4x-6
(2x+4)(4x-6)
F.O.I.L.= (8x^2-12x+16x-24)
=8x^2+4x-24
Finding a function of another function:
(fog)(x)=
f(g(x))
f(x)= 2x+4
g(x)= x-4
plug in the g(x) equation into the f(x) equation where ALL x's appear.
2(x-4)+4
distribute.
2x-8+4
(fog)(x)=2x-12
Second Late reflection.
As I stated in the last reflection,
i'm DEFINITELY behind.
So here's another blog!
I'm just explaining some of the stuff we learned in chapter 13 that I actually thought I understood.
* to find a term
arithmetic- tn = t1 + (n-1)d
where n=term #, t1=first term, d=what you add, tn=term you're looking for
geometric- tn = t1 x r^(n-1)
where r=what you multiply by
* to find the sum
arithmetic- sn = n(t1 + tn) / 2
geometric- sn = t1(1 - r^n) / 1-r
tn-1 = previous term
tn-2 = previous to the previous term
**Arithmetic sequences are when you have to add or subtract to get the next term. Geometric sequences are when you have to multiply to get the next term.
Examples:Find the formula for this arithmetic sequence:
3, 5, 7,...
tn = 3 + (n-1)2
tn = 3 + 2n - 2
tn = 1 + 2n
*These are easy you just plug in the previous term to get your answer.
You add 2 to each number to get the next number so plug 2 in for (n-1)
Solve, then you'll get the answer.
Find the next four numbers in the sequence:
t1 = 1 & tn = 3(tn-1) - 11, 2, 5, 14, 41,...
*For these you just keep plugging in the previous term. Start with t1 then go to
t4 plugging in the previous term for (tn-1)
Find the sum of the first ten terms of the series:
2 - 6 + 18 - 54 +...
s10 = 2(1 - (-3)^10) / 1 - (-3)
s10 = -29, 524
*(2 being the first number in the problem, -3 being what you multiply each number by
to get the next term)
Find the sum of the first 25 terms of the arithmetic series:
11 + 14 + 17 + 20 +...
tn = 11 +(25 - 1)3
tn = 83
s25 = 25(11 +83) / 2
s25 = 1175
*(11 being the first number in the sequence, 3 being the number you add, Plug 25 into the
n-1 formula because your looking for the 25th term)
i'm DEFINITELY behind.
So here's another blog!
I'm just explaining some of the stuff we learned in chapter 13 that I actually thought I understood.
* to find a term
arithmetic- tn = t1 + (n-1)d
where n=term #, t1=first term, d=what you add, tn=term you're looking for
geometric- tn = t1 x r^(n-1)
where r=what you multiply by
* to find the sum
arithmetic- sn = n(t1 + tn) / 2
geometric- sn = t1(1 - r^n) / 1-r
tn-1 = previous term
tn-2 = previous to the previous term
**Arithmetic sequences are when you have to add or subtract to get the next term. Geometric sequences are when you have to multiply to get the next term.
Examples:Find the formula for this arithmetic sequence:
3, 5, 7,...
tn = 3 + (n-1)2
tn = 3 + 2n - 2
tn = 1 + 2n
*These are easy you just plug in the previous term to get your answer.
You add 2 to each number to get the next number so plug 2 in for (n-1)
Solve, then you'll get the answer.
Find the next four numbers in the sequence:
t1 = 1 & tn = 3(tn-1) - 11, 2, 5, 14, 41,...
*For these you just keep plugging in the previous term. Start with t1 then go to
t4 plugging in the previous term for (tn-1)
Find the sum of the first ten terms of the series:
2 - 6 + 18 - 54 +...
s10 = 2(1 - (-3)^10) / 1 - (-3)
s10 = -29, 524
*(2 being the first number in the problem, -3 being what you multiply each number by
to get the next term)
Find the sum of the first 25 terms of the arithmetic series:
11 + 14 + 17 + 20 +...
tn = 11 +(25 - 1)3
tn = 83
s25 = 25(11 +83) / 2
s25 = 1175
*(11 being the first number in the sequence, 3 being the number you add, Plug 25 into the
n-1 formula because your looking for the 25th term)
One of the MANY late reflections.
So, obviously I'm definitely behind on this blog thing...
I have NO idea what I've been doing! On a sunday, i know i have to do them then i wake up Monday morning and think "oops"...
So here's one of four:
I'll discuss some of the formulas and trig functions we've learned recently.
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
something that i understood the most was section 2
here's some examples:
tan α = 2 and tan β=1
find tan (α - β)
= tan α + tan β/1-tan α tan β
=2+1/1-(2)(6)
=3/-1
=-3
Find the exact value of: tan 15+tan 30/1-tan 15 tan 30
tan α = 2 and tan β=1
find tan (α - β)
= tan (15 + 30)
=tan (45)
=1
Oh, and did I mention that I failed my trig exam MISERABLY? :/
& i actually studied!
I have NO idea what I've been doing! On a sunday, i know i have to do them then i wake up Monday morning and think "oops"...
So here's one of four:
I'll discuss some of the formulas and trig functions we've learned recently.
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
something that i understood the most was section 2
here's some examples:
tan α = 2 and tan β=1
find tan (α - β)
= tan α + tan β/1-tan α tan β
=2+1/1-(2)(6)
=3/-1
=-3
Find the exact value of: tan 15+tan 30/1-tan 15 tan 30
tan α = 2 and tan β=1
find tan (α - β)
= tan (15 + 30)
=tan (45)
=1
Oh, and did I mention that I failed my trig exam MISERABLY? :/
& i actually studied!
Monday, May 10, 2010
5/9
Here is how you change the base of a log.
Change of Base:
Steps:
1. Take the log (base what you want of both sides)
2. Write as an exponential
3. Move exponent to the front
4. Solve for variable
5. Write as a fraction or whole number if possible. If not possible leave in log form
Ex: 1. 5^x = 10
2. log 5^x = log 10
3. xlog 5 = 1
4. Answer: x = 1/log 5
Change of Base:
Steps:
1. Take the log (base what you want of both sides)
2. Write as an exponential
3. Move exponent to the front
4. Solve for variable
5. Write as a fraction or whole number if possible. If not possible leave in log form
Ex: 1. 5^x = 10
2. log 5^x = log 10
3. xlog 5 = 1
4. Answer: x = 1/log 5
Sunday, May 9, 2010
Last Reflection
logb^x=a
all you would do is switch the x and the a. (Kinda like an inverse)
b^a=x <---This is exponential form.
log3^2=7
3^7=2
This helps you solve problems sometimes.
But before you solve you have to change it to exponential form.
Example:
logx^4=2
x^2=4
x= +/- 2
For natural logs-
e^x=75
change e to ln(natural log), then switch the x and the 75
ln75=x
Plug into your calculator.
ln 75=4.317
all you would do is switch the x and the a. (Kinda like an inverse)
b^a=x <---This is exponential form.
log3^2=7
3^7=2
This helps you solve problems sometimes.
But before you solve you have to change it to exponential form.
Example:
logx^4=2
x^2=4
x= +/- 2
For natural logs-
e^x=75
change e to ln(natural log), then switch the x and the 75
ln75=x
Plug into your calculator.
ln 75=4.317
reflection
This week was pretty fun, dressing up in different clothes everyday for pi week. It went by kind of slow..It started off with the trig exam on monday and tuesday, which i thought i did pretty good on, and come to find out i did to pretty good:)..Then for the rest of the week we just worked on our chapter tests for the final exam in a few weeks. The tests are chapters 1-6 and 13, if anyone didn't know that. On this reflection im going to review parabolas.....
REMEMBER:
1) x=-b/2a axis of symmetry
2) 2 ways to find vertex
(-b/2a, f(-b/2a)) or comeplete the square to get vertex form.
y=(x+z+^2 +b a&b are numbers
(-a,b)---vertex
3) focus:
1/up = coeff. of x^2
then add p
*if opens up add to y-value from vertex
*opens down, subtract
*opens right add to x-value
*opens left subtract
4) directrix is p units behind vertex
always x= or y=
so subtract p
**directrix is a line, focus is a point!
**if opens up subtract & is opens down add; from y-value of vertex
**opens right subtract x-value
**opens left add x-value
EXAMPLES:
1) x^2 +1
vertex: x=-b/2a = 0/2(1) = 0
(0,1)
focus: 1/4p = 1/1 p=1/4
(0,1+1/4) = (0,5/4)
D: y=1-1/4
y=3/4
REMEMBER:
1) x=-b/2a axis of symmetry
2) 2 ways to find vertex
(-b/2a, f(-b/2a)) or comeplete the square to get vertex form.
y=(x+z+^2 +b a&b are numbers
(-a,b)---vertex
3) focus:
1/up = coeff. of x^2
then add p
*if opens up add to y-value from vertex
*opens down, subtract
*opens right add to x-value
*opens left subtract
4) directrix is p units behind vertex
always x= or y=
so subtract p
**directrix is a line, focus is a point!
**if opens up subtract & is opens down add; from y-value of vertex
**opens right subtract x-value
**opens left add x-value
EXAMPLES:
1) x^2 +1
vertex: x=-b/2a = 0/2(1) = 0
(0,1)
focus: 1/4p = 1/1 p=1/4
(0,1+1/4) = (0,5/4)
D: y=1-1/4
y=3/4
Reflection
I'll explain finding the symmetry and domain and range since that will be on our final exam.
finding the symmetry:
EXAMPLE: y=x^3=4x
a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No
b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No
c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No
d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes
Domain and Range:
These are domain and range problems with fractions and polynomials.
EXAMPLE: y=x^3+4x^2+12
For 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^2-4
First, set the bottom of the fraction equal to zero.
x^2-4=0
+4+4
x^2=4
x=+ or - 2
Your answer then comes to,
Domain: (-oo,-2)u(-2,2)u(2,oo)
finding the symmetry:
EXAMPLE: y=x^3=4x
a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No
b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No
c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No
d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes
Domain and Range:
These are domain and range problems with fractions and polynomials.
EXAMPLE: y=x^3+4x^2+12
For 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^2-4
First, set the bottom of the fraction equal to zero.
x^2-4=0
+4+4
x^2=4
x=+ or - 2
Your answer then comes to,
Domain: (-oo,-2)u(-2,2)u(2,oo)
reflection 4/9
WOOOOOOOOOOOOOOO! 3 more weeks of mathh! which means 3 more weeks of blogg. which means a whole nother year of calculus:( For people who did bad doing trig, here are some following trig stufffffffffffffff...
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
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
see everyone tomorrow!!!!
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
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
see everyone tomorrow!!!!
reflection may 9
This week was pretty fun, dressing up in different clothes everyday for pi week. It went by kind of slow..It started off with the trig exam on monday and tuesday, which i thought i did pretty good on, and come to find out i did to pretty good:)..Then for the rest of the week we just worked on our chapter tests for the final exam in a few weeks. The tests are chapters 1-6 and 13, if anyone didn't know that. On this reflection im going to review parabolas.....
REMEMBER:
1) x=-b/2a axis of symmetry
2) 2 ways to find vertex
(-b/2a, f(-b/2a)) or comeplete the square to get vertex form.
y=(x+z+^2 +b a&b are numbers
(-a,b)---vertex
3) focus:
1/up = coeff. of x^2
then add p
*if opens up add to y-value from vertex
*opens down, subtract
*opens right add to x-value
*opens left subtract
4) directrix is p units behind vertex
always x= or y=
so subtract p
**directrix is a line, focus is a point!
**if opens up subtract & is opens down add; from y-value of vertex
**opens right subtract x-value
**opens left add x-value
EXAMPLES:
1) x^2 +1
vertex: x=-b/2a = 0/2(1) = 0
(0,1)
focus: 1/4p = 1/1 p=1/4
(0,1+1/4) = (0,5/4)
D: y=1-1/4
y=3/4
REMEMBER:
1) x=-b/2a axis of symmetry
2) 2 ways to find vertex
(-b/2a, f(-b/2a)) or comeplete the square to get vertex form.
y=(x+z+^2 +b a&b are numbers
(-a,b)---vertex
3) focus:
1/up = coeff. of x^2
then add p
*if opens up add to y-value from vertex
*opens down, subtract
*opens right add to x-value
*opens left subtract
4) directrix is p units behind vertex
always x= or y=
so subtract p
**directrix is a line, focus is a point!
**if opens up subtract & is opens down add; from y-value of vertex
**opens right subtract x-value
**opens left add x-value
EXAMPLES:
1) x^2 +1
vertex: x=-b/2a = 0/2(1) = 0
(0,1)
focus: 1/4p = 1/1 p=1/4
(0,1+1/4) = (0,5/4)
D: y=1-1/4
y=3/4
Reflection 5/9
Okay, so I liked this week overall, besides the fact that we took the two trig exams, which were fairly easy and i got a B so i'm happy. Happy Mothers' Day to all you mothers out there and B-Rob of course. Now, I'm going to discuss some limit stuff we learned a couple of chapters back.
Limits are denoted as lim with and n (arrow) infinity underneath it.
Here are the rules:
1. lim(n infinity) - if the degree of the top = the degree of the bottom then the answer is the coefficients
2. If the degree of the top is greater than the degree of the bottom, then it is infinity
3. If the degree of the top is smaller than the degree of the bottom, then it is 0.
IF the rules don't apply, you will have to use your calculator to find what the sequence is approaching.
For a geometric sequence if the absolute value of r is less than 1, then it goes to 0.
Examples:
lim (n infinity) sin (1/n)
sin(1/100) = .010
sin(1/1000) = .0010
sin(1/10000) = .00010
lim (n infinity) n+5/n = 1, because the degree is the same, so the coefficients equal 1
I'm also having some trouble with trying to simplify equations that deal with the trig functions. I guess I just need to learn my formulas more and know how to apply them to different types of equations and problems. After all, I'm going to know how to do this 100x more next year for Calculus next year, I'm guessing. So can some people give me some examples on how to do the simplifying with these problems? Thanks.
Limits are denoted as lim with and n (arrow) infinity underneath it.
Here are the rules:
1. lim(n infinity) - if the degree of the top = the degree of the bottom then the answer is the coefficients
2. If the degree of the top is greater than the degree of the bottom, then it is infinity
3. If the degree of the top is smaller than the degree of the bottom, then it is 0.
IF the rules don't apply, you will have to use your calculator to find what the sequence is approaching.
For a geometric sequence if the absolute value of r is less than 1, then it goes to 0.
Examples:
lim (n infinity) sin (1/n)
sin(1/100) = .010
sin(1/1000) = .0010
sin(1/10000) = .00010
lim (n infinity) n+5/n = 1, because the degree is the same, so the coefficients equal 1
I'm also having some trouble with trying to simplify equations that deal with the trig functions. I guess I just need to learn my formulas more and know how to apply them to different types of equations and problems. After all, I'm going to know how to do this 100x more next year for Calculus next year, I'm guessing. So can some people give me some examples on how to do the simplifying with these problems? Thanks.
Reflection
This week we took our 2 trig test and worked on review packets for other sections. I find chapter 11 is pretty simple. I understood 11-3 and 11-4. We learned stuff about cardioids, limacons, rose, petals, archimedes spirals, and logmarithmisc spirals. The quiz on that was easy because its simple to understand the formulas with the graphs.
EXAMPLES FROM 11-4 AND 11-3
1)re^i(theta)=rcis(theta)
3e^i(2pi)=3cis2pi
=3cos2pi+3sin2pi(i)
=3(1)+3(0)i
=3
2)De Moivre's Thereom
z^n=r^ncis(n)(theta)
z=2cis20degrees Find z^2
z^2=2^2cis2(20degrees)
z^2=4cis40degrees
3)De Moivre's Thereom
4cis15degrees Find z^4
z^4=4^4cis4(15degrees)
z^4=256cis60degrees
4)Find z^1/4, z=16cis180degrees
z^1/4=16^1/4cis(180*/4+0X360*/4)= 2cis45*
z^1/4=16^1/4cis(180*/4+1X360*/4)= 2cis135*
z^1/4=16^1/4cis(180*/4+2X360*/4)= 2cis225*
z^1/4=16^1/4cis(180*/4+3X360*/4)= 2cis315*
5)Find z^1/2, z=9cis60*
z^1/2=9^1/2cis(60*/2+0X360*)= 3cis30*
z^1/2=9^1/2cis(60*/2+1X360*)= 3cis210*
EXAMPLES FROM 11-4 AND 11-3
1)re^i(theta)=rcis(theta)
3e^i(2pi)=3cis2pi
=3cos2pi+3sin2pi(i)
=3(1)+3(0)i
=3
2)De Moivre's Thereom
z^n=r^ncis(n)(theta)
z=2cis20degrees Find z^2
z^2=2^2cis2(20degrees)
z^2=4cis40degrees
3)De Moivre's Thereom
4cis15degrees Find z^4
z^4=4^4cis4(15degrees)
z^4=256cis60degrees
4)Find z^1/4, z=16cis180degrees
z^1/4=16^1/4cis(180*/4+0X360*/4)= 2cis45*
z^1/4=16^1/4cis(180*/4+1X360*/4)= 2cis135*
z^1/4=16^1/4cis(180*/4+2X360*/4)= 2cis225*
z^1/4=16^1/4cis(180*/4+3X360*/4)= 2cis315*
5)Find z^1/2, z=9cis60*
z^1/2=9^1/2cis(60*/2+0X360*)= 3cis30*
z^1/2=9^1/2cis(60*/2+1X360*)= 3cis210*
REFLECTION 5/9
Only a few more blogs left you guysss!! I'm soooo excited haha. Well anyway, this week we started the review for our exam, which is Chapters 1 through 6 and Chapter 13. I'm just going to go over a few things from Chapter 4 this week. So here's a few examples:
Ex. 1) -2x^2 + 3x - 6 Find the vertex.
*To find the vertex all you have to use is the axis of symmetry formula which is:
x = -b/2a and you plug in what you get for that for the y coordinate
In other words, this is what the vertex would be:
(-b/2a,f(-b/2a))
*So when you plug into -b/2a you get: x = -3/2(-2) So x = 3/4
Now you plug 3/4 into the equation to get the y coordinate
So you get: -2(3/4)^2 + 3(3/4) - 6
= -9/8 + 9/4 - 6
= -39/8
**So your vertex is (3/4,-39/8)
Ex. 2) y = x-2x Find the inverse and prove that it's an inverse.
*Okay first of all, you have to type this function into your calculator to see the graph. And if the graph passes the horizontal line test (the line only touches the graph at one point) then the function has an inverse.
*The graph is a line so it has an inverse
*Now to find the inverse you take the equation you're given, switch the x's and y's and solve for y.
*So you get: x = y-2y
x=-y
So the inverse is y = -x
Prove:
1. F(f^-1(x)) = (-x) - 2(-x)
= -x+2x
=x (*if you get "x" as the answer, then you know you're right)
2. F^-1(f(x)) = -(x-2x)
=-x+2x
=x
Ex. 3) f(x)=x+6 g(x)=3x+1 **Find g(f(x))
*All this function is telling you to do is to plug the f equation into the g equation every time you see an x
*So you get 3(x+6)+1
=3x+18+1
*So your answer is 3x+19
(:
Ex. 1) -2x^2 + 3x - 6 Find the vertex.
*To find the vertex all you have to use is the axis of symmetry formula which is:
x = -b/2a and you plug in what you get for that for the y coordinate
In other words, this is what the vertex would be:
(-b/2a,f(-b/2a))
*So when you plug into -b/2a you get: x = -3/2(-2) So x = 3/4
Now you plug 3/4 into the equation to get the y coordinate
So you get: -2(3/4)^2 + 3(3/4) - 6
= -9/8 + 9/4 - 6
= -39/8
**So your vertex is (3/4,-39/8)
Ex. 2) y = x-2x Find the inverse and prove that it's an inverse.
*Okay first of all, you have to type this function into your calculator to see the graph. And if the graph passes the horizontal line test (the line only touches the graph at one point) then the function has an inverse.
*The graph is a line so it has an inverse
*Now to find the inverse you take the equation you're given, switch the x's and y's and solve for y.
*So you get: x = y-2y
x=-y
So the inverse is y = -x
Prove:
1. F(f^-1(x)) = (-x) - 2(-x)
= -x+2x
=x (*if you get "x" as the answer, then you know you're right)
2. F^-1(f(x)) = -(x-2x)
=-x+2x
=x
Ex. 3) f(x)=x+6 g(x)=3x+1 **Find g(f(x))
*All this function is telling you to do is to plug the f equation into the g equation every time you see an x
*So you get 3(x+6)+1
=3x+18+1
*So your answer is 3x+19
(:
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