8-5
This section was probably the hardest thing we’ve learned all year, at least that’s what I think. In 8-5 we use identities to get to the same trig function.
Things you can’t do to solve:
1. Divide by a trig function when solving to cancel.
2. Cancel from the inside of a trig function.
EXAMPLE 1: 2 sin^2 theta -1 = 0
A). We can tell by the sin^2 that we are going to have to take an inverse to solve this problem.
B). First, move the 1 to the right. Then divide by just the 2 from the left and right.
C). Now, take the square root of the left and right sides – square root sin^2 theta = square root ½
D). sin theta = +/- square root ½
E). According to our trig chart, ½ is 45 degrees. Since it is a square root, find all the coordinates.
F). sin theta = 45 degrees, 135 degrees, 225 degrees, 315 degrees
Hope everyone is having a fun, relaxing holiday! Doing these blogs only reminds me that we go back to school MONDAY, oh joy..
Tuesday, December 28, 2010
Holiday..
Alright, so I think this is my third post of the holidays (not prompts).
Going back to Chapter 10 now..
Chapter 10 dealt with finding the exact value of cos, sin, and tan. In 10-1, we were given two formulas:
cos(alpha +/- beta) = cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta) = cos alpha cos beta +/- cos apha cos sin beta
EXAMPLE 1: Show that sin (3pi/2 – x) = -cos
A). We see the left side of our sin formula here.. so lets expand it – sin 3pi/2 cos x - cox 3pi/2 sin x
B). Replace the 3pi/2 with numbers from your trig chart. *3pi/2 = 270 degrees
C). (-1) cos x - (0) sin x
D). - cos x = - cos x
EXAMPLE 2: Solve cos (90 degrees + theta) + cos (90 degrees - theta)
A). Expand the left and right side of the (+) – cos 90 degrees cos theta - sin 90 degrees sin theta + cos 90 degrees cos theta + sin 90 degrees sin theta
B). The -/+ sin 90 degrees sin theta cancels out and you’re left with cos 90 degrees cos theta + cos 90 degrees cos theta
C). 2(cos 90 degrees)(cos theta)
D). Plug in cos 90 degrees from your trig chart or unit circle.
E). 2(0)cos theta
F). Your answer is cos theta
Going back to Chapter 10 now..
Chapter 10 dealt with finding the exact value of cos, sin, and tan. In 10-1, we were given two formulas:
cos(alpha +/- beta) = cos alpha cos beta -/+ sin alpha sin beta
sin(alpha +/- beta) = cos alpha cos beta +/- cos apha cos sin beta
EXAMPLE 1: Show that sin (3pi/2 – x) = -cos
A). We see the left side of our sin formula here.. so lets expand it – sin 3pi/2 cos x - cox 3pi/2 sin x
B). Replace the 3pi/2 with numbers from your trig chart. *3pi/2 = 270 degrees
C). (-1) cos x - (0) sin x
D). - cos x = - cos x
EXAMPLE 2: Solve cos (90 degrees + theta) + cos (90 degrees - theta)
A). Expand the left and right side of the (+) – cos 90 degrees cos theta - sin 90 degrees sin theta + cos 90 degrees cos theta + sin 90 degrees sin theta
B). The -/+ sin 90 degrees sin theta cancels out and you’re left with cos 90 degrees cos theta + cos 90 degrees cos theta
C). 2(cos 90 degrees)(cos theta)
D). Plug in cos 90 degrees from your trig chart or unit circle.
E). 2(0)cos theta
F). Your answer is cos theta
Monday, December 27, 2010
Chapter 10 Flashback
Going back to Chapter 10 because I know I did not remember this to well for the exam. You couldn't use calculators for most of this chapter, but you really didn't need them if you knew your trig chart. This chapter only had 4 sections in it, and out of them the first section was definately the easiest. There were only two formulas to learn. It was pretty basic math, and calculators were still not allowed. Still had to know the trig chart if you had any intentions of finishing a problem.
The two formulas were for sin and cos:
Sin ( alpha + beta ) = ( sinalpha ) ( cosbeta ) + ( cosalpha ) (sinbeta)
Cos ( alpha + beta ) = ( cosalpha ) ( cosbeta ) – ( sinalpha ) (sinbeta )
Example:Find the exact value of cos 150
Cos ( 90 + 60 ) = cos90 cos60 – sin90 sin60(0)(1/2) – (1)(square root 3/3)
0 – square root 3/3
Answer: Square root3/3
** If you don't understand where the 0, 1/2, 1, and square root of 3/3 is coming from.. it's coming from you sin and cos of the degrees on the trig chart.
The two formulas were for sin and cos:
Sin ( alpha + beta ) = ( sinalpha ) ( cosbeta ) + ( cosalpha ) (sinbeta)
Cos ( alpha + beta ) = ( cosalpha ) ( cosbeta ) – ( sinalpha ) (sinbeta )
Example:Find the exact value of cos 150
Cos ( 90 + 60 ) = cos90 cos60 – sin90 sin60(0)(1/2) – (1)(square root 3/3)
0 – square root 3/3
Answer: Square root3/3
** If you don't understand where the 0, 1/2, 1, and square root of 3/3 is coming from.. it's coming from you sin and cos of the degrees on the trig chart.
Chapter 9 Review
I am going back to Chapter 9 because these types of problems are often found on the ACT andI know I sometimes struggle with them. In Chapter 9, section 1 we learn how to solve right triangles. This lesson is a simple repeat of what we had learned in geometry. To help solve these problems and make memorizing the formulas easier you must remember SOHCAHTOA.
SOHCAHTOA is:
Sin (S)= opposite (O)/hypotenuse (H)
Cos (C)= adjacent (A)/hypotenuse (H)
Tan (T)= opposite (O)/adjacent (A)
Once you can remember SOHCAHTOA, you can remember all the formulas. Secant is the reciprocal of sine, cosecant is the reciprocal of cosine, and cotangent is the reciprocal of tangent.When looking at a triangle, opposite is the degrees across from the angle already given to you. The degrees already attached to the given is the adjacent angle, and hypotenuse is ALWAYS the degrees across from the right angle.
To solve a problem, draw your triangle and plug in the points and angles given to you. Some triangles may be named. Once you have drawn the triangle, find what they are asking you to solve (the missing angles or points), and you look for the appropriate formula. Once you have a degree to each angle you are done. Make sure to label all angles and points on the triangle, if you want credit.
Solve the triangle QRS.
Angle R- 34degrees Side r- 26 degrees Angle Q- 90 degrees
Those are the angles given, so you need to find angle and sides q, S, and s.
In order to find S, simply subtract 180-90-34 to compose an answer.
Angle S= 56 degrees
In order to find q, use the formula for sin which is opposite/ hypotenuse
Angle q= 46.496 degrees
In order to find s, use the formula for tan, which is opposite/ adjacent
Angle s= 38.547 degrees
SOHCAHTOA is:
Sin (S)= opposite (O)/hypotenuse (H)
Cos (C)= adjacent (A)/hypotenuse (H)
Tan (T)= opposite (O)/adjacent (A)
Once you can remember SOHCAHTOA, you can remember all the formulas. Secant is the reciprocal of sine, cosecant is the reciprocal of cosine, and cotangent is the reciprocal of tangent.When looking at a triangle, opposite is the degrees across from the angle already given to you. The degrees already attached to the given is the adjacent angle, and hypotenuse is ALWAYS the degrees across from the right angle.
To solve a problem, draw your triangle and plug in the points and angles given to you. Some triangles may be named. Once you have drawn the triangle, find what they are asking you to solve (the missing angles or points), and you look for the appropriate formula. Once you have a degree to each angle you are done. Make sure to label all angles and points on the triangle, if you want credit.
Solve the triangle QRS.
Angle R- 34degrees Side r- 26 degrees Angle Q- 90 degrees
Those are the angles given, so you need to find angle and sides q, S, and s.
In order to find S, simply subtract 180-90-34 to compose an answer.
Angle S= 56 degrees
In order to find q, use the formula for sin which is opposite/ hypotenuse
Angle q= 46.496 degrees
In order to find s, use the formula for tan, which is opposite/ adjacent
Angle s= 38.547 degrees
Exam Review
This week has been a review for exams and what not. We've been retaking our tests as practice for our exams/midterms. As I was taking my tests I realized that I was having trouble with chapter thirteen :/ So I will Blog on one of those sections. Other than chapter thirteen, and maybe chapter ten, i think i did pretty well.
Chapter thirteen is on sequences and series.
a sequence is a list of numbers.
a series is a list of numbers being added together.
sequences are either geometric or arithmetic.
the term t(n) simply means term number.. term term2 etc.
The problems i had the most difficulty with were the ones in which you had to solve for tn given a t(b0 or an t(c)
EX.
find t10 for t3=7 and t7=15
first find the formula.
tn=2(n)+1
t10=2(10)+1
t10=21
after you find the formula, the problem becomes a lot easier.
Chapter thirteen is on sequences and series.
a sequence is a list of numbers.
a series is a list of numbers being added together.
sequences are either geometric or arithmetic.
the term t(n) simply means term number.. term term2 etc.
The problems i had the most difficulty with were the ones in which you had to solve for tn given a t(b0 or an t(c)
EX.
find t10 for t3=7 and t7=15
first find the formula.
tn=2(n)+1
t10=2(10)+1
t10=21
after you find the formula, the problem becomes a lot easier.
Tuesday, December 21, 2010
11-3
11-3
Throwing back to Chapter 11.. Section 11-3 was very short and simple. It dealt with De Moivre’s Theorem. There’s just one simple formula for this section:
(r cis theta)^n = r^n cis n theta
EXAMPLE 1: If z=3 cis 10^6, find z^6.
A). z^6 = 3^6 cis 6(10)
B). z^6 = 729 cis 60
Pretty simple right?
EXAMPLE 2: If z=2 cis 4^3, find z^3.
A). z^3 = 2^3 cis 3(4)
B). z^3 = 8 cis 12
This is too short so I’m going to add an example problem from another section of Chapter 11.
EXAMPLE 1: Convert to rectangular – (3, pi)
A). We should all remember our formulas but if not, here they are..
x= r cos theta and y=r sin theta
B). x=3 cos pi and y=3 cos pi
C). According to our unit circle, we know that pi is 180. At 180 degrees cos is -1 (x axis)and sin is 0 (y axis).
D). x=3(-1) and y=3(0)
E). Your answer is the point (-3,0)
Throwing back to Chapter 11.. Section 11-3 was very short and simple. It dealt with De Moivre’s Theorem. There’s just one simple formula for this section:
(r cis theta)^n = r^n cis n theta
EXAMPLE 1: If z=3 cis 10^6, find z^6.
A). z^6 = 3^6 cis 6(10)
B). z^6 = 729 cis 60
Pretty simple right?
EXAMPLE 2: If z=2 cis 4^3, find z^3.
A). z^3 = 2^3 cis 3(4)
B). z^3 = 8 cis 12
This is too short so I’m going to add an example problem from another section of Chapter 11.
EXAMPLE 1: Convert to rectangular – (3, pi)
A). We should all remember our formulas but if not, here they are..
x= r cos theta and y=r sin theta
B). x=3 cos pi and y=3 cos pi
C). According to our unit circle, we know that pi is 180. At 180 degrees cos is -1 (x axis)and sin is 0 (y axis).
D). x=3(-1) and y=3(0)
E). Your answer is the point (-3,0)
Thursday, December 16, 2010
Now, my last post of the holidays, I will be doing law of sin like I said I would in the blog before this one.
In law of sin, you need angles opposite of each other. (not in a right triangle)
Example: a = 45, A = 30, B = 75 Find b.
The formula for law of sin is: sin ( angle 1 ) / opp. Angle X sin ( angle 2 )/ opp. Angle
Plug it in. sin 30/45 X sin 75/b
b sin 30 = 45 sin 75
b= 45 sin 75 / sin 30 ( plug into your calculator exactly as you see it. ) * also make sure you are in degree mode.
b = 86.9
I finished all my blogs, yay!! So, I hope everyone has a MERRY CHRISTMAS and HAPPY NEW YEARS!! (:
( and Kaitlyn, if you’re reading this, HAPPY BIRTHDAY (: incase you don’t get my text like last year haha)
In law of sin, you need angles opposite of each other. (not in a right triangle)
Example: a = 45, A = 30, B = 75 Find b.
The formula for law of sin is: sin ( angle 1 ) / opp. Angle X sin ( angle 2 )/ opp. Angle
Plug it in. sin 30/45 X sin 75/b
b sin 30 = 45 sin 75
b= 45 sin 75 / sin 30 ( plug into your calculator exactly as you see it. ) * also make sure you are in degree mode.
b = 86.9
I finished all my blogs, yay!! So, I hope everyone has a MERRY CHRISTMAS and HAPPY NEW YEARS!! (:
( and Kaitlyn, if you’re reading this, HAPPY BIRTHDAY (: incase you don’t get my text like last year haha)
Now it is time for blog #2. Another thing I remember well is chapter nine. In chapter 9, we used SOHCAHTOA to find missing angles of RIGHT triangles. Also, the law of sin was pretty easy too. So, I guess in this blog I will show you how to do SOHCAHTOA and in the next I will do law of sin.
Okay, starting off with SOHCAHTOA. *Which can only be used on right triangles.
Example: a = 56, A = 25, C = 90 Find angles B and b
Since C = 90, that means you have a right triangle. It is asking you to find the two b’s. Finding B is very simple because all triangles will equal 180 degrees. All you have to do is subtract 180 from angles A and C.
B = 65
Now to find little b, we will be using tan, because that is the simplest formula to use in this problem. We are looking for b, which is opposite and we already have the adjacent angle which is 56.
Tan65 = b/56
b= 56tan65 (plug into calculator) * make sure you are in degree mode
b= 120
Okay, starting off with SOHCAHTOA. *Which can only be used on right triangles.
Example: a = 56, A = 25, C = 90 Find angles B and b
Since C = 90, that means you have a right triangle. It is asking you to find the two b’s. Finding B is very simple because all triangles will equal 180 degrees. All you have to do is subtract 180 from angles A and C.
B = 65
Now to find little b, we will be using tan, because that is the simplest formula to use in this problem. We are looking for b, which is opposite and we already have the adjacent angle which is 56.
Tan65 = b/56
b= 56tan65 (plug into calculator) * make sure you are in degree mode
b= 120
Okay so, since I woke up super early today I decided to do all of my math blogs to get them out the way so I don’t have to worry about math until we get back to school. But like a dummy, I left my binder at school with allll of our notes and everything in it, so I’ll be doing simple problems of things I fully remember how to do. And also since we haven’t learned anything new recently, I’m not exactly sure what we’re supposed to blog about? But anyways I’ll go back to chapter 7 and go over coterminal angles and converting radians to degrees.
A coterminal angle is simply adding or subtracting 360 to a number.
Find a positive coterminal angle to 225.
1. Since it is positive, you will be adding 360. Answer: 585 degrees
Simple as that!
Convert 4pi to degrees.
1. 4pi/1 X 180/pi
2. The two pi’s cancel out, and when you cross multiply, you are left with 720degrees
Answer: 720 degrees
A coterminal angle is simply adding or subtracting 360 to a number.
Find a positive coterminal angle to 225.
1. Since it is positive, you will be adding 360. Answer: 585 degrees
Simple as that!
Convert 4pi to degrees.
1. 4pi/1 X 180/pi
2. The two pi’s cancel out, and when you cross multiply, you are left with 720degrees
Answer: 720 degrees
Sunday, December 12, 2010
This past week we did lotss of study guides to help review for our mid- term exams that are coming up. We got back all of our old tests and had to redo them. I saw that i did not understand some of the old things that we did earlier in the year, so i learned it. All the test were from chapters 7 through 13. In chapter 7, we learned things about degrees and radians and how to switch back and forth to each. Each chapter we re-applied all of the knowledge that we learned the chapter before. Know all your formulas, and you will pass the test. I don’t know all my formulas, so i might not pass the test. I’m reallyyy nervous for this exam. The section that i am having the most problems with is the identities. I just can’t remember the formulas for some reason. I just don’t like chapter 8 at all. But i think i can learn it by the time the exam comes.
I’m like everyone else, since we didn’t learn anything new this week I’m not sure what to blog about. Hopefully I’m not doing this wrong. I’m so excited that Christmas holidays are almost here and we FINALLY get time off to relax. Although, I forgot a lot of things we learned in math over Thanksgiving, so who knows what’s going to happen over 3 weeks! It took a lot of hard work to keep a B in this class, and I am hoping I can do the same thing for the rest of the year. I just have to pay attention as much as I can. When I took the ACT again this weekend, it had more geometry and algebra II than ever. I felt like I couldn’t remember a lot of formulas for things, so I don’t feel like I did too well on the math section. Hopefully I’ll be surprised, but I did know how to do most of the advanced math questions. Yay!
Mid Term Exam
This week in advanced math we reviewed for our Mid-Term Exam. We did everything from chapter 7 to chapter 13. From changing degrees to radians to solving arithmetic and geometric equations. We came a long way and now we are getting our knowledge put to the test. My favorite chapter through out these sections is of course chapter 7. My least favorite chapter is chapter 13. The section I improved the most in is 8-3 identities. The section I still need work on that is not chapter 13 is chapter 9. I have learned a lot this semester and I believe it will show on my ACT. I knew way more than I knew before I started taking advanced math. For now though here is an example of my favorite problem.
Ex. Convert 15Ï€/7 to degrees
15π/7* 180/π = 2700π/7π=385.7º
Midterm review
This week is review week for the MIDTERM exam. We were given chapter 7-13 old test to work for extra help and points. Basically, we didn't do nothing new and I'm going talk about how I feel for this midterm exam and how u can bang bang pass it.
Basically if you know all the formulas and the concept of each sections. This midterm exam should be pretty long and easy. The only way I feel that it can be hard if you give up and do not know any formulas or any idea of how to work the problems. This test is big and it is like a big percentage.
Know your formulas and concept of each sections. Ex: law of cosine, law of sine, rad to degrees, degrees to rad, and basically everything else ha.
Example problem from chapter 7
446 44` 20`=446.7390
Basically if you know all the formulas and the concept of each sections. This midterm exam should be pretty long and easy. The only way I feel that it can be hard if you give up and do not know any formulas or any idea of how to work the problems. This test is big and it is like a big percentage.
Know your formulas and concept of each sections. Ex: law of cosine, law of sine, rad to degrees, degrees to rad, and basically everything else ha.
Example problem from chapter 7
446 44` 20`=446.7390
Ehhh.
Its time for a new exciting installment of Wren's Sunday Blog. This week we will be discussing varying topics including our upcomin exam. First on the list, ACT. Whoever took it knows what was on it. Personally I had the most trouble with Geometry. I just couldn't remember any of the formulas. Next on our list of topics, is why parents are such terrible people. I believe its because they want us to be unhappy. I know they have our best interests at heart but is it worth having your child like you less and less everytime we talk. When it comes to school parents need to gently remind us of things, not make threats. My dad is threatening to make me take all my exams in order to pull up some grades. If he does this I'm considering bombing everyone of them as revenge. Finally, our math exam. I am not looking forward to this. Not because I'm bad at math but because i am to lazy to study for it. So because of this laziness I will probably make a bad grade and feel the wrath of my parents. Great now my moms yelling at me for other stuff. I'm, moving out. Not really, but if I could I would. Peace
This past week in advanced math we reviewed for our midterm exam. If you are exempt (unlike me) we will take our exam this Friday. I am extremely nervous because I absolutely suck at math. I try as hard as can but I still come up short when it comes to getting that B. I have got to come up with some kind of way to get it. Anyway our exam will consist of everything we have done this semester. That means chapters, 7, 8, 9, 10, 11, and 13 will be one there. I’m pretty good with chapters 7 and 11 and some bits on 9, 10, and 13 but when it comes to chapter 8 I’m pretty much screwed. Identities are not my thing. The only good thing about these upcoming exams is that I should be exempt from all my other exams so that means I can spend my whole week studying math and after that we get two weeks off!!!!!!!! (: YAYYYYYY! Sorry if I didn’t do this blog right. We didn’t learn anything new so I guess I’ll just through in a random example problem from chapter 7.
Random Example:
Convert 95° to radians
95° x Ï€/180 = 19Ï€/180
Super easy fun stuff right there! (:
Random Example:
Convert 95° to radians
95° x Ï€/180 = 19Ï€/180
Super easy fun stuff right there! (:
Sooo this past week all we did was retake our old tests for an exam review. Well all those test I thought I did good on apparently I didn't. I honestly had trouble with some of them but when I asked you a question I remembered all over again :)
I'm not really sure what to blog about since we haven't learned anything new so I figured i'll just talk about how I feel about math. I don't like it. I think math hates me as much as I hate it but i'm trying really hard to do well in advanced math and so far i'm surprising myself. I'm not doing half as bad as I thought i was considering how bad I thought algebra 2 was. It's a constant struggle to make sure I fully understand everything so I prepare myself for tests. Doing homework helps a lot but I never know if i'm doing the work right because we don't get the answers to check them. My #1 goal is to pass this exam and i'm terrified!
I'm not really sure what to blog about since we haven't learned anything new so I figured i'll just talk about how I feel about math. I don't like it. I think math hates me as much as I hate it but i'm trying really hard to do well in advanced math and so far i'm surprising myself. I'm not doing half as bad as I thought i was considering how bad I thought algebra 2 was. It's a constant struggle to make sure I fully understand everything so I prepare myself for tests. Doing homework helps a lot but I never know if i'm doing the work right because we don't get the answers to check them. My #1 goal is to pass this exam and i'm terrified!
An Old Concept For A New Exam
This past week we reviewed for our midterms. In Advanced Math, it means relearning everything you thought you escaped from in the first nine weeks. Being the forgetful person I am, this was like cramming everything from the first day of school back into my head. But this did help me realize that I do need to do more studying. So, I decided for this blog to go back and reteach myself chapter nine. In this chapter, we learned how to use SOHCAHTOA to find different sides of a triangle.
SOHCAHTOA stands for Sin equals Opposite over Hypotenuse, Cos equals Adjacent over Hypotenuse, Tan equals Opposite over Adjacent. Besides those three, Csc, Sec, and Cot are also used. Csc equals hypotenuse over opposite Sec equals hypotenuse over adjacent and Cot equals adjacent over opposite.
But, there are some restrictions. These can only be used with a right triangle. The hypotenuse is the longest side which means it is opposite of the right angle. Never confuse the hypotenuse with the adjacent side that would be a bad idea.
Here's an example-
If you look out of a third story window 20 feet in the air to the top of a skyscraper 400 feet away and the angle of elevation is 35 degrees you and the top of the skyscraper, how tall is the skyscraper?
In this case, you would take tan 35 and set it equal to y over 400.
Then solve for Y.
Y = 400 tan 35
Y then equals 280.083.
After that you must add 20 because of the 20 feet you were already up in the air.
Your total would then be 300.083
SOHCAHTOA stands for Sin equals Opposite over Hypotenuse, Cos equals Adjacent over Hypotenuse, Tan equals Opposite over Adjacent. Besides those three, Csc, Sec, and Cot are also used. Csc equals hypotenuse over opposite Sec equals hypotenuse over adjacent and Cot equals adjacent over opposite.
But, there are some restrictions. These can only be used with a right triangle. The hypotenuse is the longest side which means it is opposite of the right angle. Never confuse the hypotenuse with the adjacent side that would be a bad idea.
Here's an example-
If you look out of a third story window 20 feet in the air to the top of a skyscraper 400 feet away and the angle of elevation is 35 degrees you and the top of the skyscraper, how tall is the skyscraper?
In this case, you would take tan 35 and set it equal to y over 400.
Then solve for Y.
Y = 400 tan 35
Y then equals 280.083.
After that you must add 20 because of the 20 feet you were already up in the air.
Your total would then be 300.083
Holiday Blog Prompt 3
What concept(s) from Algebra II did you most struggle with? Why? We will be revisiting many Algebra II concepts this semester, what do you plan to do differently to master them?
Holiday Blog Prompt 2
Explain the different types of polar graphs and their equations. Find sites with images of each. **Hint use google and search the images tab. Images can be pasted into blogger and include the link. You should have different sites.
Holiday Blog Prompt 1
Come up with your own trig graphing problem and walk someone through it step by step explaining each step and formula as though they have never taken Advanced Math.
Saturday, December 11, 2010
13-3
Section 13-3 was about finding the sum of a series. There are two formulas – one for geometric and one for arithmetic.
The sum of the first n terms of an arithmetic series:
Sn=n(t1+tn)/2
The sum of the first n terms of a geometric series:
Sn=t1(1-r^n)/1-r
Along with this section, there were a few terms we learned.
series: a list of numbers being added together
finite: a certain number
infinite: unlimited number of terms
EXAMPLE 1: Find the sum of the first 20 terms of the arithmetic series:
12 + 15 + 18 + 21 + 24..
A). This is an arithmetic series so we’re going to use the first formula.
B). Plug in your information given: S20=20(12 + t20)/2
C). We don’t know what t20 is, so we have find it by going back to what we learned in 13-1.
D). tn=12 + (20-1)3
E). By solving you find that t20=69. Now we can continue step B.
F). S20=20(12 + 69)/2
G). S20=810
EXAMPLE 2: How many multiples of 7 are there between 10 and 70?
A). Start off by figuring out the first few multiples: 7, 14, 21, 28.. and find the last one which is 63.
B). Now go back to 13-1 and use the formula tn=t1+(n-1)d
C). 63=7+(n-1)7
D). n=9
The sum of the first n terms of an arithmetic series:
Sn=n(t1+tn)/2
The sum of the first n terms of a geometric series:
Sn=t1(1-r^n)/1-r
Along with this section, there were a few terms we learned.
series: a list of numbers being added together
finite: a certain number
infinite: unlimited number of terms
EXAMPLE 1: Find the sum of the first 20 terms of the arithmetic series:
12 + 15 + 18 + 21 + 24..
A). This is an arithmetic series so we’re going to use the first formula.
B). Plug in your information given: S20=20(12 + t20)/2
C). We don’t know what t20 is, so we have find it by going back to what we learned in 13-1.
D). tn=12 + (20-1)3
E). By solving you find that t20=69. Now we can continue step B.
F). S20=20(12 + 69)/2
G). S20=810
EXAMPLE 2: How many multiples of 7 are there between 10 and 70?
A). Start off by figuring out the first few multiples: 7, 14, 21, 28.. and find the last one which is 63.
B). Now go back to 13-1 and use the formula tn=t1+(n-1)d
C). 63=7+(n-1)7
D). n=9
Monday, December 6, 2010
Week 7 Prompt
What have you personally accomplished in math this semester? Do you feel more confident? What will you do differently next semester?
Sunday, December 5, 2010
Smh
This had to be by far the hardest section for me to grasp. I don't know what happened with the test but I didn't understand it. But I guess the show must go on. In chapter 13 we learned lots of things, mostly based around geometric and arithmetic sequences. In 13-1 we learned the arithmetic sequence, geometric sequence, and neither. To find the arithmetic sequence you use the formula tn= t1 + (n-1)d. To find the geometric sequence you use the formula tn= t1(r^(n-1)). And it goes on and on from there with complicated things. But the one thing I did get was Sigma.
Sigma
#
∑f(x) ; the # signs are the limits of sumation. The n is the index. And f(x) is the summand.
n=#
Ex.
6
∑ 4x
x=1
1) What is the summand? 4x
2) What is the index? x
3) What are the limits of summation? 6 and 1
4) Evaluate the sigma
4+4+4+4+4+4=24
(4*1)+(4*2)+(4*3) +(4*4)+(4*5)+4*6)= 84
13-3
We just took a test on the whole chapter 13. This chapter was challenging and pretty complex. The chapter I had easy with was chapter 13. Chapter 13 section 13-3 is dealing with arithmetic series and the sum of the 1st n term of it. The formula for this is sn=n(t1+tn)/2
The sum of the 1st n terms of a geometric series is sn=t1(1-r^n)/1-r
Series- a list of numbers being added together
Finite – is a certain number
Infinite-unlimited number of terms
Example one arithmetic and one geometric problem:
Find the sum of the 1st 25 terms of the arithmetic series.
11+14+17+20+
tn=11+(25-1)3=83
s25=25(11+83)/2=1175 is your answer
Find the sum of the 1st 10 terms of the geometric series of 2-6+18-54+
-3
S10=2(1-3)^10/1-3=-29,524 is your answer
If you know your formulas and rules and everything .Section 13-3 is really easy and not hard and you should breeze by it easy. Know your definitions and formulas and everything and u will be good.
The sum of the 1st n terms of a geometric series is sn=t1(1-r^n)/1-r
Series- a list of numbers being added together
Finite – is a certain number
Infinite-unlimited number of terms
Example one arithmetic and one geometric problem:
Find the sum of the 1st 25 terms of the arithmetic series.
11+14+17+20+
tn=11+(25-1)3=83
s25=25(11+83)/2=1175 is your answer
Find the sum of the 1st 10 terms of the geometric series of 2-6+18-54+
-3
S10=2(1-3)^10/1-3=-29,524 is your answer
If you know your formulas and rules and everything .Section 13-3 is really easy and not hard and you should breeze by it easy. Know your definitions and formulas and everything and u will be good.
Chapter 13
Chapter 13 has been a rough one for me! I know i could have drilled it more into my brain but with a week off, I began to slack! So i appologize for that. It's been a rough week, and honestly I would really like to get my credit for doing this blog, but somehow i left my binder in your classroom. This week we've been doing our study guides for our exam, and i feel confident in our previous Chapters, especially Chapter 7. I feel I have Chapter 7 down pack. When we return to school we will continue our study guides, and hopefully not get yelled at for our horrible grades on Chapter 13's test grades because I know a lot of us didn't do to well on them.
RIP Taylor Adams.
and Sorry once again Mrs. Robinson for this no good blog, but at least i posted.
RIP Taylor Adams.
and Sorry once again Mrs. Robinson for this no good blog, but at least i posted.
Bombed it
Yeah I bombed that test. I bombed hard. Just about every answer was an educated guess. Personally i blame Thanksgiving. Since most of our chapter was learned and then we had a week off I forgot eveything. Some may say that i had a whole week to relearn everthing but its not that easy. And since everyone was looking forward to the break and no one was paying attention in class its not our fault. I think before exams we need to have an extensive review of everything. Being a junior sucks.
Alright, so i thought it would be appropriate for me to actually do my blog BUT i forgot my binder at school. I would just like to say that by reading trey perrin's blog i think he is kind of confident that he did good on his chapter 13 test. So,i would like to know his grade when we get them back. And also that Brooke's blog looks pretty cool.
Okay, so we just finished chapter 13 and to be honest i pretty much forgot it all once i started the Chapter 7 and 8 review packets. Also, that when you werent there on friday i slept during the test because i couldn't think of anything we learned. So when you see the test you know why. Well, what i do remember is that we learned about sequences (arithmetic and geometric) and that the formulas are-
Sum of arithmetic series:
Sn=n(t1+tn) /2
Sum of geometric series:
Sn=t1(1-r^n)/1-r
Also, we learned how to figure out series or sequences. We also learned about sigmas and what to do with them, and recursive definitions. (I'd tell you if i had my binder)
That is pretty much what i know. Sorry about not having any examples, but hey 1 point is better than a 0.
Okay, so we just finished chapter 13 and to be honest i pretty much forgot it all once i started the Chapter 7 and 8 review packets. Also, that when you werent there on friday i slept during the test because i couldn't think of anything we learned. So when you see the test you know why. Well, what i do remember is that we learned about sequences (arithmetic and geometric) and that the formulas are-
Sum of arithmetic series:
Sn=n(t1+tn) /2
Sum of geometric series:
Sn=t1(1-r^n)/1-r
Also, we learned how to figure out series or sequences. We also learned about sigmas and what to do with them, and recursive definitions. (I'd tell you if i had my binder)
That is pretty much what i know. Sorry about not having any examples, but hey 1 point is better than a 0.
13-4
Sooo.... I am one of the many who believes that they did not do well on their test. I completely forgot that section thirteen - four even existed and I think this is what made me bomb it.
Alright, here we go.
Rules for fractions.
If the top degrees equals the bottom degrees then the answer is coefficient.
If the top degrees is bigger than the bottom degrees, the answer is plus or minus infinity.
If the top degree is less than the bottom degrees, the answer is zero.
If those rules don't apply, use a table and figure out what it is approaching.
lim r^n = 0 if |r| < 1
r = number n = infinity
Example lim
n = infinity = (.99)^n = 0
|.99| < 1
Alright, here we go.
Rules for fractions.
If the top degrees equals the bottom degrees then the answer is coefficient.
If the top degrees is bigger than the bottom degrees, the answer is plus or minus infinity.
If the top degree is less than the bottom degrees, the answer is zero.
If those rules don't apply, use a table and figure out what it is approaching.
lim r^n = 0 if |r| < 1
r = number n = infinity
Example lim
n = infinity = (.99)^n = 0
|.99| < 1
Sooo just like everyone else except TREY PERRIN, I failed that chapter 13 test. I actually thought i knew chapter 13 until I got that test so I guess i'm kinda upset. But what's a better way to learn than to blog about it! yayyyy, so i'll talk about 13-5 since i found that one not as bad.
it's all about the sum of an infinite geometric.
FORMULA:
sn=t1/1-r
writing a repeating decimal as a fraction is my favorite!
FORMULA:
#thats repeating/last place-1
EXAMPLES:
find the sum of the infinite geometric series.
9-6+4-....
*plug into your formula
sn=9/1-2/3
your answer is 27/5
write .2525 as a fraction
*plug into formula
25/100-1
your answer is 25/99
it's all about the sum of an infinite geometric.
FORMULA:
sn=t1/1-r
writing a repeating decimal as a fraction is my favorite!
FORMULA:
#thats repeating/last place-1
EXAMPLES:
find the sum of the infinite geometric series.
9-6+4-....
*plug into your formula
sn=9/1-2/3
your answer is 27/5
write .2525 as a fraction
*plug into formula
25/100-1
your answer is 25/99
The only thing I can remember I learned these past few weeks is a ARITHMETIC SEQUENCES :D
Sequence is just an order or pattern. For Example: *&(*&(*&( The consecutive sequence is *&(
Arithmetic is basically just using +, -, x , or / (basic mathematics)
Arithmetic Sequence, is just an order or pattern using basic mathematics.
Otays... When dealing with Arithmetic Sequences, you aren't going to have consecutive numbers, you are going to have different numbers with a consecutive *beat* flowing throughout the problem... So you have a set of numbers put into a particular order; lets say...... 2 5 8 11 ... otay.. this sequence can go on and on. BUT thats not the point.. Our obligation is to find out how to get the numbers by finding out what is making these numbers go up? so we look to see whats the amount of numbers in between (btw. in between should be ONE word)... 2 and 5 --- 3 numbers in between~~~5 and 8 --- 3 numbers in between~~~ 8 and 11 --- 3 numbers in between...so we found our arithmetic pattern.. YAHHH!
13-1
This week in advanced math we did not learn anything new. We reviewed for our chapter 13 test which we took Friday. We also received our chapter 7 and chapter 8 packets for our exam next week. So let’s do a little review of chapter 13. Section 13-1 is introducing sequences. Sequences are a list of numbers. There are two kinds of sequences, arithmetic and geometric. An arithmetic sequence is where the same number is added or subtracted each time. Geometric sequences are where the same number is multiplied or divided each time. Just like every other chapter there are formulas involved. Both arithmetic and geometric sequences have their own formulas:
Arithmetic - tn = t1 + (n-1) d
Geometric – tn = t1 = rn-1
Tn – the term
Examples:
Is the following sequences arithmetic, geometric, or neither?
3, 6, 9, 12, 15, 18, 21 – Arithmetic d = 3
5, 7, 10, 13, 16, 22, 24 – Neither random numbers
2, 8, 32, 128, 512, 2048 – Geometric d = 4
Arithmetic - tn = t1 + (n-1) d
Geometric – tn = t1 = rn-1
Tn – the term
Examples:
Is the following sequences arithmetic, geometric, or neither?
3, 6, 9, 12, 15, 18, 21 – Arithmetic d = 3
5, 7, 10, 13, 16, 22, 24 – Neither random numbers
2, 8, 32, 128, 512, 2048 – Geometric d = 4
thirteen.two
This week is our last week before exams :D I don't know about you, but that's good news to me :) two weeks ago we finished chapter thirteen excluding section seven. this past week we have been reviewing for exams and Friday we took our 13 test: hopefully I did well :D
Anyways, one of the sections tht I didn't find too tricky was section thirteen two.
section thirteen two is on Recursive Definitions. Recursive Definitions are sequences that are defined by what came before.
ex. t(n-1) means the number before. t(n-2) means two numbers before and so on.
example: 6,11,16,21..
we know that 5 is being added consistently each time.Therefore, the sequence is an arithmetic one. the formula is
tn=t(n-1) + 5 .
ex. find t3 if t1=5 t2=10 and t(n)= 2t(n-1) + t(n-2)
2(10)+5 = 25
t3 is 25.
simple enough :)
Anyways, one of the sections tht I didn't find too tricky was section thirteen two.
section thirteen two is on Recursive Definitions. Recursive Definitions are sequences that are defined by what came before.
ex. t(n-1) means the number before. t(n-2) means two numbers before and so on.
example: 6,11,16,21..
we know that 5 is being added consistently each time.Therefore, the sequence is an arithmetic one. the formula is
tn=t(n-1) + 5 .
ex. find t3 if t1=5 t2=10 and t(n)= 2t(n-1) + t(n-2)
2(10)+5 = 25
t3 is 25.
simple enough :)
December 5
Unlike everyone else, I think I actually did well on my chapter 13 test. I am confident because I have the BEST advanced math teacher in the whole world. That test wasnt too bad. Anyways, this week in class we took a chapter 13 test and also we did our previous test for extra practice to get us ready for our advanced math exam. I am very confident going into this exam because I feel like I have retained a lot of information over the first two nine weeks. I like how my teacher gives us our previous test back to work as a study guide so we can be sure that we have remembered what we learned and over the first half of the school year. Also there is no reason we should be failing this course because she stays after school just about every day for the kids who are struggling or need help in the class. I think that she is preparing us the best that she can for college and she is the best teacher in the world. :)
13-1
Chapter 13 was the last chapter we have learned in class. We just took a VERY hard chapter test on it before the weekend. It was all about sequences, series, and sigmas. The formulas must be memorized just like any other type of math problem in order to do well. I was going to do this blog on 13-6 with the sigmas but I can’t figure out how to get a picture of a sigma on here, so I’ll do 13-1. 13-1 was pretty simple because you needed to know arithmetic and geometric sequences.
Aritmetic- when a number is added or subtracted repeatedly
Geometric- when a number is multiplied or divided repeatedly
Ex. 2, 7, 12, 17, 22… is an arithmetic sequence because 5 was added each time
1, 3, 9, 27… is geometric because 3 was multiplied to each number
There are formulas for both as well, tn = t1 + (n-1) (d) – arithmetic
tn = t1 x (r) ^ n-1 – geometric
Those formulas are used for different things such as finding which number would come later on in a series and different things like that.
Aritmetic- when a number is added or subtracted repeatedly
Geometric- when a number is multiplied or divided repeatedly
Ex. 2, 7, 12, 17, 22… is an arithmetic sequence because 5 was added each time
1, 3, 9, 27… is geometric because 3 was multiplied to each number
There are formulas for both as well, tn = t1 + (n-1) (d) – arithmetic
tn = t1 x (r) ^ n-1 – geometric
Those formulas are used for different things such as finding which number would come later on in a series and different things like that.
Saturday, December 4, 2010
13-6
So, who else failed that Chapter test? HAHA. Looks like everyone needs another holiday already..
Section 6 of 13 was about sigmas. A sigma is a series written in condensed form. If you didn’t know, a series is a list of numbers being added together. This section was fairly easy if you knew the parts of the sigma.
EXAMPLE 1:
6
∑ 3k
k = 1
Limits of summation – 1, 6
Index – k
Summand – 3k
A). Expand the sigma.
3(1) + 3(2) + 3(3) + 3(4) + 3(5) + 3(6) + 3(7)
3 + 6 + 9 + 12 + 15 + 18 + 21
B). Evaluate the sigma.
3 + 6 + 9 + 12 + 15 + 18 + 21 = 78
EXAMPLE 2: Express using a sigma – 7 + 3 -1 -5 -9
5
∑ 11-4k
k = 1
*k=1 when working backwards!
Section 6 of 13 was about sigmas. A sigma is a series written in condensed form. If you didn’t know, a series is a list of numbers being added together. This section was fairly easy if you knew the parts of the sigma.
EXAMPLE 1:
6
∑ 3k
k = 1
Limits of summation – 1, 6
Index – k
Summand – 3k
A). Expand the sigma.
3(1) + 3(2) + 3(3) + 3(4) + 3(5) + 3(6) + 3(7)
3 + 6 + 9 + 12 + 15 + 18 + 21
B). Evaluate the sigma.
3 + 6 + 9 + 12 + 15 + 18 + 21 = 78
EXAMPLE 2: Express using a sigma – 7 + 3 -1 -5 -9
5
∑ 11-4k
k = 1
*k=1 when working backwards!
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