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 15π/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 π/180 = 19π/180
Super easy fun stuff right there! (:
Random Example:
Convert 95° to radians
95° x π/180 = 19π/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!
Monday, November 29, 2010
13-6
13-6 is all about sigmas. A sigma is simply a series written in condensed form. Since I'm not fancy with my computer I'm going to use an E to represent a sigma and I will try to express the rest as best as I can.
On the top of the sigma is the limit of summation (#). Underneath the sigma is where the index is placed (n). And to the right is the summand (f(x)).
Example
7
E 2k
k=1
Here the limits are 7 and 1
The summand is 2k
The index is k
In order to evaluate the sigma replace k with the numbers 1-7 like so
2*1+2*2+2*3+2*4+2*5+2*6+2*7
simplify
2+4+6+8+10+12+14
Then evaluate and you get 56
Now, expand
8
E 2i(^2)-1
i=5
2*5(^2)-1+2*6(^2)-1+2*7(^2)-1+2*8(^2)-1
In expanded form this ^ is 49+71+97+127
Simple enough, right?
On the top of the sigma is the limit of summation (#). Underneath the sigma is where the index is placed (n). And to the right is the summand (f(x)).
Example
7
E 2k
k=1
Here the limits are 7 and 1
The summand is 2k
The index is k
In order to evaluate the sigma replace k with the numbers 1-7 like so
2*1+2*2+2*3+2*4+2*5+2*6+2*7
simplify
2+4+6+8+10+12+14
Then evaluate and you get 56
Now, expand
8
E 2i(^2)-1
i=5
2*5(^2)-1+2*6(^2)-1+2*7(^2)-1+2*8(^2)-1
In expanded form this ^ is 49+71+97+127
Simple enough, right?
Week 6 Prompt
How do you determine if a sequence is arithmetic or geometric? What are the rules for finding limits? Give examples of each.
Sunday, November 28, 2010
Braindead
The title says it all, I am braindead. The only reason I even remember what Chapter we are on is because I read other people posts. So what I gathered from that is we are on Chapter 13. Chapter thirteen is about sequence and series and all this other stuff. I think we have exams coming up soon, oh joy. I need to do really good on these exams so if anybody is thinking about having study groups or something let me know. I hope everyone had a Happy Thanksgiving. I ate a lot of food. Then we lost to John Curtis officially ending the 2010 football season. I'm bored now. Goodnight.
chapter 13- section 3
Happy late Turkey Day everyone! Lucky for us, we get to go back to school tomorrow! We were taught all Sections of Chapter 13 before the holidays, and its the last chapter before our exam review for the next few weeks. This wasn't the easiest chapter for me, and ill need some review when we get back, because i have a lot of unclear problems and a lot of confusion. Section 13-3 was the most simple of the whole chapter because it was mostly just finding the answers from the guildlines in your notes.
Guidlines for this Chapter:
For Fractions:
1. If the top degree = bottom degree the answer is a coefficient.
2. If the top degree > bottom degree the answer is +/- infinity.
3. If the top degree < bottom degree the answer is 0.
If the rules above do not apply to the given, then you simply use the table function in your calculator.
For example:
1. Lim/n->infinity n^7 +6/ 9n^2 – 7n
The degrees are 7 and 5.
The top one (7) is larger than the bottom one (2).
When you look at your rules, they state that the answer will be +/- infinity.
see you tomorrow (:
Guidlines for this Chapter:
For Fractions:
1. If the top degree = bottom degree the answer is a coefficient.
2. If the top degree > bottom degree the answer is +/- infinity.
3. If the top degree < bottom degree the answer is 0.
If the rules above do not apply to the given, then you simply use the table function in your calculator.
For example:
1. Lim/n->infinity n^7 +6/ 9n^2 – 7n
The degrees are 7 and 5.
The top one (7) is larger than the bottom one (2).
When you look at your rules, they state that the answer will be +/- infinity.
see you tomorrow (:
Before the holidays we did chapter 13. 13-2 was about recursive- defines the terms of what came before.
FORMULAS:
to find:
previous term---tn-1
two terms back---tn-2
1)find the certain number of terms.
2)give a recursive definition.
EXAMPLES:
find the 3rd, 4th, and 5th terms.
tn=4tn-8
t1=10
just multiply the number in front of t and the first terms and subtract 8.
t2=4(10)-8 = 32
t3=4(32)-8 = 120
t4=4(120)-8 = 472
t5=4(472)-8 = 1,880
It's very simple to find terms like this especially when you are given the 1st term already.
The other way is to find a recursive definition.
Give a recursive definition for 2,4,6,8....
**you can obviously see that 2 is being added each time.
*ARITHMETIC PATTERN!
RECURSIVE DEFINITION:
tn=tn-2+2
SEE ALL OF YOU TOMORROW :)
FORMULAS:
to find:
previous term---tn-1
two terms back---tn-2
1)find the certain number of terms.
2)give a recursive definition.
EXAMPLES:
find the 3rd, 4th, and 5th terms.
tn=4tn-8
t1=10
just multiply the number in front of t and the first terms and subtract 8.
t2=4(10)-8 = 32
t3=4(32)-8 = 120
t4=4(120)-8 = 472
t5=4(472)-8 = 1,880
It's very simple to find terms like this especially when you are given the 1st term already.
The other way is to find a recursive definition.
Give a recursive definition for 2,4,6,8....
**you can obviously see that 2 is being added each time.
*ARITHMETIC PATTERN!
RECURSIVE DEFINITION:
tn=tn-2+2
SEE ALL OF YOU TOMORROW :)
13-4
I hope everyone had a great Thanksgiving holidays. But now that we have school tomorrow it is time to get your mind focus to school again. Section 13-4 is basically the rules for fractions and how to apply it.
Rule for fractions are:
-If top degree = bottom degree; answer is coeff
-If top degree > bottom degree; answer is +/- infinity
-If top degree < bottom degree; answer is 0
If rules don’t apply you use a table and figure out what it is approaching.
Lim
n->infinity r^n=0 iff absolute value r<1
Examples:
Lim n->infinity(.99)^n=0
absolute value .99^n<1
.366,..,.0004,.000….2=0 is your answer
lim n->infinity n^2+1/2n^2-3n=1/2
lim n->infinity sim(1/n)=0
lim n->infinity 5n^2+square root n/3n^3+7=0
lim n->infinity 7n^3/4n^2-5=infinity
Basically section 13-4 is really easy and common sense. If you know your rules and know how to apply them and use them you will not get any trouble at all in section 13-4 and this section should be a piece of cake for you.
Rule for fractions are:
-If top degree = bottom degree; answer is coeff
-If top degree > bottom degree; answer is +/- infinity
-If top degree < bottom degree; answer is 0
If rules don’t apply you use a table and figure out what it is approaching.
Lim
n->infinity r^n=0 iff absolute value r<1
Examples:
Lim n->infinity(.99)^n=0
absolute value .99^n<1
.366,..,.0004,.000….2=0 is your answer
lim n->infinity n^2+1/2n^2-3n=1/2
lim n->infinity sim(1/n)=0
lim n->infinity 5n^2+square root n/3n^3+7=0
lim n->infinity 7n^3/4n^2-5=infinity
Basically section 13-4 is really easy and common sense. If you know your rules and know how to apply them and use them you will not get any trouble at all in section 13-4 and this section should be a piece of cake for you.
13-2
Recursive Definitions( Recursive means define interms of what came before)
tn-1/previous terms/ tn=tn-1 - 3
tn-2/two terms back
tn-3/three less than preceding term
Ex:1 tn=4tn-1 +1 and t1=7 find the 3,4,5 terms.
t2=4(7) + 1=29
t3=4(29)+1=117
t4=4(117)+1=469
t5=4(469)+1=1877
Ex:2 Give a recursive definiton for 7,10,13,16..
- since the pattern is +3, the definition would be tn=tn-1 + 3
Arithmetic and Geometric Series and their Sums
I hope everyone had a wonderful break :) I know the past week off was appreciated, and for me, needed. But anyways back to business...
Section thirteen three is basically on the sums of the geometric and arithmetic series. This section, like the rest of the chapter is relatively simple. Some important terms to remember are Finite, which is a certain number, and infinite, which is an unlimited number of terms. There are two formulas; one for each sequence and you simply replace, or substitute the numbers given. Once you have learned to solve these types of problems, it will be easier to solve the more difficult ones.
Sum of arithmetic series :
Sn=n(t1+tn) /2
Sum or geometric series:
Sn=t1(1-r^n)/1-r
Where r is the common ratio and is not equal to one.
Ex. Find the sum of the arithmetic series .
1. S10: t1=3,t10=39
S10=10(3+39)/2
S10=10(42)/2
S10=420/2
S10=210
Ex. Find the sum of the geometric series.
1. Find the sum if the first 50 terms.
2,4,8..
• Because the t50, or the fiftieth term is not given you must go back to section one to find it.
T50=2+(50-1)2
T50=2+(49)2
T50=100
S50=2(1-2^50)/1-50
Finally you just solve to get your answer.
:)
Section thirteen three is basically on the sums of the geometric and arithmetic series. This section, like the rest of the chapter is relatively simple. Some important terms to remember are Finite, which is a certain number, and infinite, which is an unlimited number of terms. There are two formulas; one for each sequence and you simply replace, or substitute the numbers given. Once you have learned to solve these types of problems, it will be easier to solve the more difficult ones.
Sum of arithmetic series :
Sn=n(t1+tn) /2
Sum or geometric series:
Sn=t1(1-r^n)/1-r
Where r is the common ratio and is not equal to one.
Ex. Find the sum of the arithmetic series .
1. S10: t1=3,t10=39
S10=10(3+39)/2
S10=10(42)/2
S10=420/2
S10=210
Ex. Find the sum of the geometric series.
1. Find the sum if the first 50 terms.
2,4,8..
• Because the t50, or the fiftieth term is not given you must go back to section one to find it.
T50=2+(50-1)2
T50=2+(49)2
T50=100
S50=2(1-2^50)/1-50
Finally you just solve to get your answer.
:)
13-2
Recursive Definitions
Section 13-2 was the second easiest section in chapter 13. The term recursive means to define in the terms of what came before.
Previous term – tn-1
Two terms back – tn-2
….. – tn-3
The two main directions you will follow for this section is to give a recursive definition for a series of numbers and finding a certain number of terms. Let’s try some examples:
Example 1:
Find the 3rd, 4th, 5th, and 6th terms
Tn = 2tn-1 + 7 and t1 = 3
T2 = 2(3) + 7 = 13
T3 = 2(13) + 7 = 33
T4 = 2(33) + 7 = 73
T5 = 2(73) +7 = 153
T6 =2(153) + 7 = 313
Example 2:
Give a recursive definition for
2, 6, 10, 14
Tn = tn-1 + 4
10, 20, 40, 80
Tn = tn-1(2)
Hope everyone had a great Thanksgiving holiday! Now only 2 more weeks and 1 week of exams (unless your exempt) until Christmas holidays!
Section 13-2 was the second easiest section in chapter 13. The term recursive means to define in the terms of what came before.
Previous term – tn-1
Two terms back – tn-2
….. – tn-3
The two main directions you will follow for this section is to give a recursive definition for a series of numbers and finding a certain number of terms. Let’s try some examples:
Example 1:
Find the 3rd, 4th, 5th, and 6th terms
Tn = 2tn-1 + 7 and t1 = 3
T2 = 2(3) + 7 = 13
T3 = 2(13) + 7 = 33
T4 = 2(33) + 7 = 73
T5 = 2(73) +7 = 153
T6 =2(153) + 7 = 313
Example 2:
Give a recursive definition for
2, 6, 10, 14
Tn = tn-1 + 4
10, 20, 40, 80
Tn = tn-1(2)
Hope everyone had a great Thanksgiving holiday! Now only 2 more weeks and 1 week of exams (unless your exempt) until Christmas holidays!
13-4
I hope everyone had a great Thanksgiving holidays! But unfortunately, we have to go back to a few more weeks of school before Christmas. Before the holidays, we learned chapter 13, which is the last chapter before our exam. This chapter was pretty difficult for me, so I will need a good review when we go back to school tomorrow. Section 13-4 was sort of simple because MOST of the time there were answers given by looking at your notes. The notes we were given are:
For Fractions:
1. If top degree = bottom degree; answer is coeff
2. If top degree > bottom degree; answer is +/- infinity
3. If top degree < bottom degree; answer is 0
If none of these rules apply, the you use the table in your calculator to find out which number it is approaching.
For example:
1. Lim/n->infinity n^5 +4/ 6n^2 – 7n
The degrees are 2 and 5. The top one (2) is smaller than the bottom one (5).
When you look at your rules, they state that the answer will be 0.
That is your final answer, so these types of problems are pretty simple.
For Fractions:
1. If top degree = bottom degree; answer is coeff
2. If top degree > bottom degree; answer is +/- infinity
3. If top degree < bottom degree; answer is 0
If none of these rules apply, the you use the table in your calculator to find out which number it is approaching.
For example:
1. Lim/n->infinity n^5 +4/ 6n^2 – 7n
The degrees are 2 and 5. The top one (2) is smaller than the bottom one (5).
When you look at your rules, they state that the answer will be 0.
That is your final answer, so these types of problems are pretty simple.
Saturday, November 27, 2010
13-2
13-2
Section 13-2 was about recursive definitions. Recursive means define in terms of what came before.
tn-1: previous term
tn-2: two terms back
tn-3: …
EXAMPLE 1: tn=3tn-1 + 1 and tn=6. Find the 3, 4, and 5 terms.
A). t2=3(6) + 1 = 19
t3=3(19) + 1 = 58
t4=3(58) + 1 = 175
t5=3(175) + 1 = 526
B). 58, 175, and 526 are your 3, 4, and 5 terms. All you do is replace tn-1 with tn to begin. Then you take your answer and place is where tn-1 is each time.
EXAMPLE 2: Give a recursive definition for 6, 10, 14, 16..
A). Looking at the numbers above, you can easily see that you’re adding 4 each time to make this arithmetic pattern.
B). Your recursive definition is tn=tn-1 +4
Hope everyone had a great, relaxing Thanksgiving break :)
Section 13-2 was about recursive definitions. Recursive means define in terms of what came before.
tn-1: previous term
tn-2: two terms back
tn-3: …
EXAMPLE 1: tn=3tn-1 + 1 and tn=6. Find the 3, 4, and 5 terms.
A). t2=3(6) + 1 = 19
t3=3(19) + 1 = 58
t4=3(58) + 1 = 175
t5=3(175) + 1 = 526
B). 58, 175, and 526 are your 3, 4, and 5 terms. All you do is replace tn-1 with tn to begin. Then you take your answer and place is where tn-1 is each time.
EXAMPLE 2: Give a recursive definition for 6, 10, 14, 16..
A). Looking at the numbers above, you can easily see that you’re adding 4 each time to make this arithmetic pattern.
B). Your recursive definition is tn=tn-1 +4
Hope everyone had a great, relaxing Thanksgiving break :)
Wednesday, November 24, 2010
Week 6 Prompt
Due to the fact that the internet was not dependable on my laptop I was unable to post the blog prompt for the holidays. Therefore it is a freebie for everyone. Just make sure to post your 2 regular blogs. They can come from any review topic.
Sunday, November 21, 2010
13-6
This week in advanced math we learned chapter 13. In 13-1 we learned geometric and arithmetic sequences. In Chapter 13-2 we learned Recursive Definitions. In chapter 13-3 We learned Arithmetic and Geometric Series and their sums. In chapter 13-4 we learned Limits of Infinite Sequences. In chapter 13-5 we learned Sums of Infinite series. And finally in 13-6 we learned Sigma Notation. The greek letter sigma is often used in mathematics to express a series or its sum in abbreviated form. The Sigma contains three parts. The summand, the limits of summation, and the index. The number to the right of the Sigma is the summand. The number on top of the Sigma is the limits of summation. The number on the bottom of the Sigma is the index. You can either be asked to expand the Sigma or evaluate the sigma. To expand you only plug the numbers were they need to go. To evaluate you solve the whole thing.
Ex.Evaluate k=1; summand = (k+4); ;limit of summation = 25
1+4=5+4=9+4=13+4=17+4=21+4=25
Ex. Expand k=1; summand (k^3); limits of summation = 15
1^2 + 2^2 + 3^2... 15^2
Chapter 13.
What's up people? I hope you all get to enjoy your thanksgiving vacation. I will be spending mine practicing and I couldn't be happier. We mad it to the third round baby. I know none of ya'll believe we can win but poo on ya'll. We got a shot baby and I'm pumped. This week we started Chapter 13. This Chapter should be easy but of course the Math Gods had to screw it up. These sequences should just be a simple matter of using logic but instead it has been turnedd into an arduous process with multiple formulas to memorize. It is possible to figure out anything that has to do with these sequences by using the formulas. The easiest part of 13 was 13-6. It has to do with something called enigmas. These are a simple matter of pretty much reading a graph. Happy Thanksgiving everybody.
13-4! Uhhh.....
I am super pumped for this week off!! My brain needs a rest. Haha.
Now it's math time....
So, here are some rules for fractions
**If the top degrees equals the bottom degrees, the answer is the coefficient
**If the top degrees is greater than the bottom degrees, the answer is plus or minus infinity
**If the top degrees is less than the bottom degrees, the answer is zero
****If none of these rules apply use a table and figure out what it is approaching
-limit rn=0 and n=infinity if |r|<1 and r is a real number
I am not too sure how to do this but here is an example of how it looks like it would work.
Limit
n=infinity=(-.99)n=0
|-.99| < 1
Limit
n=infinity n2+ 1/2n2-3n = 1/2
Answers are .50766, .50075, .50008
Now it's math time....
So, here are some rules for fractions
**If the top degrees equals the bottom degrees, the answer is the coefficient
**If the top degrees is greater than the bottom degrees, the answer is plus or minus infinity
**If the top degrees is less than the bottom degrees, the answer is zero
****If none of these rules apply use a table and figure out what it is approaching
-limit rn=0 and n=infinity if |r|<1 and r is a real number
I am not too sure how to do this but here is an example of how it looks like it would work.
Limit
n=infinity=(-.99)n=0
|-.99| < 1
Limit
n=infinity n2+ 1/2n2-3n = 1/2
Answers are .50766, .50075, .50008
13-6
In advanced math this week we started chapter 13. I found that chapter 13 was very hard. However, the easiest section was section 13-6. This section is about a symbol called sigma. A sigma is a series written in condensed form. Around the sigma are its index, summand and limits of summation. The two main things you do with sigma is evaluate and expand it. Let’s try a few examples.
Example 1:
Identify
6
∑4k
K = 2
Find the Summand? 4k
What is the index? k
What are the limits of summation? 5 and 2
Evaluate the Sigma
4(2) + 4(3) + 4(4) + 4(5) + 4(6)
8 + 12 + 16+ 20 + 24 = 80
Example 2:
Expand the Sigma
10
∑ 5k4
K = 3
5(3)4 + 5(4)4 + 5(5)4 + 5(6)4 + 5(7)4 + 5(8)4 + 5(9)4 + 5(10)4
405 + 1280 + 3125 + 6480 + 12005 + 20480 + 32805 + 50000
Example 1:
Identify
6
∑4k
K = 2
Find the Summand? 4k
What is the index? k
What are the limits of summation? 5 and 2
Evaluate the Sigma
4(2) + 4(3) + 4(4) + 4(5) + 4(6)
8 + 12 + 16+ 20 + 24 = 80
Example 2:
Expand the Sigma
10
∑ 5k4
K = 3
5(3)4 + 5(4)4 + 5(5)4 + 5(6)4 + 5(7)4 + 5(8)4 + 5(9)4 + 5(10)4
405 + 1280 + 3125 + 6480 + 12005 + 20480 + 32805 + 50000
Last week we were pretty much taught all of chapter 13. It's all very easy as long as you remember your formulas.
The easiest for me was 13-4 so ill do a few problems from that.
FORMULAS:
if the degrees at the top is = to the degrees at the bottom then the answer is coefficient.
if the degrees at the top is > than the degrees at the bottom then the answer is +/- infinity.
if the degrees at the top is < than the degrees at the bottom then the answer is 0.
--if none of these rules apply then you would just plug it into your calculator using a table.
EXAMPLES:
2n^4/2n^3
**2nd rule applies!
since 4 is greater than 3 your answer is infinity.
* but is is positive or negative?!
POSITIVE INFINITY.
4n+1/4n
4n is = to 4n so you are left with 1.
your answer is 1!
HAVE A GOOD WEEK OFF EVERYONE :)
The easiest for me was 13-4 so ill do a few problems from that.
FORMULAS:
if the degrees at the top is = to the degrees at the bottom then the answer is coefficient.
if the degrees at the top is > than the degrees at the bottom then the answer is +/- infinity.
if the degrees at the top is < than the degrees at the bottom then the answer is 0.
--if none of these rules apply then you would just plug it into your calculator using a table.
EXAMPLES:
2n^4/2n^3
**2nd rule applies!
since 4 is greater than 3 your answer is infinity.
* but is is positive or negative?!
POSITIVE INFINITY.
4n+1/4n
4n is = to 4n so you are left with 1.
your answer is 1!
HAVE A GOOD WEEK OFF EVERYONE :)
13-5
Last week we were taught sections from Chapter 13. In 13-5 we learned the sum of an infinite geometric problems. 13-5 is dealing with alot of things such as the following stuff: infinite geometric and repeating decimal and infinite series converage and sequence converage.
The formula for a infinite geometric is sn=ti/1-r
When you want to write a repeating decimal as a faction you do #repeating/lastplace^-1
Examples:
-Find the sum of the infinite geometric series of 9-6+4-
-6/9=-2/3
-4/6=-2/3
Sn=9/1-(-2/3)=27/5 is the answer.
-Write .infinite5 as a faction
5/10-1=5/9 is the answer.
-For what values of x does the following infinite series converage
1+(x-2)+(x-2)^2+(x-2)^3+
x-2/1=x-2
(x-2)^2/x-2=x-2
(x-2)^3/(x-2)^2=x-2
-1
1
Section 13-5 is really easy and not hard at all. If you know the rules and the formulas and know how to do the repeating and whatever else you have to do in section 13-5. This section will be a breeze and easy for you.
The formula for a infinite geometric is sn=ti/1-r
When you want to write a repeating decimal as a faction you do #repeating/lastplace^-1
Examples:
-Find the sum of the infinite geometric series of 9-6+4-
-6/9=-2/3
-4/6=-2/3
Sn=9/1-(-2/3)=27/5 is the answer.
-Write .infinite5 as a faction
5/10-1=5/9 is the answer.
-For what values of x does the following infinite series converage
1+(x-2)+(x-2)^2+(x-2)^3+
x-2/1=x-2
(x-2)^2/x-2=x-2
(x-2)^3/(x-2)^2=x-2
-1
1
Section 13-5 is really easy and not hard at all. If you know the rules and the formulas and know how to do the repeating and whatever else you have to do in section 13-5. This section will be a breeze and easy for you.
Chapter 13- Section 6
This past week we learned Chapter 13, and completed 13-6 on friday to finish up the chapter before the holidays. We had to learn this chapter quick because we needed to finish so we could review before our midterm. For the unfortunate students who are not exempt, we will be taking a chapter test and preparing for the midterm exam that alot of us are worried about passing. The most recent section we learned is 13 – 6. Overall this chapter has been prettty simple. Please excuse my examples for this section, it is kind of hard to do without an image like we've drawn in our notebooks.
- For the problems we are solving in this sectin a “sigma” is drawn for every problem.
- A sigma may look like an E to you, but it is not.
- A sigma is a series written in a condensed (not as long) form.
- At the top of the sigma, there is a number called a limit of summation.At the bottom, there is a letter that will always equal a number.
- Just follow directions to complete the problems you are given of this type. If it asks for a particular piece of the sigma, you just find it. If it asks you to evaluate, you expand the sigma and solve it. If it tells you to expand, you just expand and stop there.
It's pretty difficult to show you exactly what to do on the computer, and without a picture, but i tried my best to sum it up.
13-6
This week we finished up learning chapter 13 before the holidays. This is the last chapter we are learning before our midterm. For the next weeks, we will be taking a chapter test and preparing for the midterm exam that we are all hoping to be exempt from. The most recent section we learned is 13 – 6. It was pretty simple, but pretty hard to do on the computer, so I will do my best.
For this section a “sigma” is drawn for every problem. A sigma looks sort of like an E. A sigma is a series written in a condensed form.
At the top of the sigma, there is a number called the limits of summation.
At the bottom, there is a letter that equals a number. Ex. K=4 The k would be the index and the number is another limits of summation.
And in the middle there is an equation called the summand.
To complete problems in this chapter all you have to do is follow directions. If it asks for a particular piece of the sigma, you just find it. If it asks you to evaluate, you expand the sigma and solve it. If it tells you to expand, you just expand and stop there. It is pretty simple, but without me being able to draw a sigma on the computer, it would be difficult for me to work any problems.
For this section a “sigma” is drawn for every problem. A sigma looks sort of like an E. A sigma is a series written in a condensed form.
At the top of the sigma, there is a number called the limits of summation.
At the bottom, there is a letter that equals a number. Ex. K=4 The k would be the index and the number is another limits of summation.
And in the middle there is an equation called the summand.
To complete problems in this chapter all you have to do is follow directions. If it asks for a particular piece of the sigma, you just find it. If it asks you to evaluate, you expand the sigma and solve it. If it tells you to expand, you just expand and stop there. It is pretty simple, but without me being able to draw a sigma on the computer, it would be difficult for me to work any problems.
13-4
13-4
Last week we were taught sections from Chapter 13. In 13-4, we were taught how to figure out what an equation’s limit by three simple rules.
1. If degrees top = degrees bottom, answer is coefficient.
2. If degrees top > degrees bottom, answer is +/- infinity.
3. If degrees top < degrees bottom, answer is 0.
*If none of these rules apply, use your calculator and figure out what it is approaching.
EXAMPLE 1: limit: n – infinity.. n^2-1/n^2
A). According to the rules, n^2 is = to n^2.
B). Therefore, take the number in front of n, which is 1 – 1/1
C). Your answer is 1.
EXAMPLE 2: limit: n – infinity.. 2n^3/ 2n^2
A). According to the rules, n^3 is > n^2
B). Therefore your answer is + infinity.
EXAMPLE 3: limit n – infinity.. log[sin(1/n)]
A). Since none of the rules apply to this problem, plug it into your calculator in the table.
B). Your answer is 0.
When plugging into your calculator, look at the table and see what the number is approaching. For instance, if it’s .01, … , .001, …, .0001 then your answer would be 0.
Last week we were taught sections from Chapter 13. In 13-4, we were taught how to figure out what an equation’s limit by three simple rules.
1. If degrees top = degrees bottom, answer is coefficient.
2. If degrees top > degrees bottom, answer is +/- infinity.
3. If degrees top < degrees bottom, answer is 0.
*If none of these rules apply, use your calculator and figure out what it is approaching.
EXAMPLE 1: limit: n – infinity.. n^2-1/n^2
A). According to the rules, n^2 is = to n^2.
B). Therefore, take the number in front of n, which is 1 – 1/1
C). Your answer is 1.
EXAMPLE 2: limit: n – infinity.. 2n^3/ 2n^2
A). According to the rules, n^3 is > n^2
B). Therefore your answer is + infinity.
EXAMPLE 3: limit n – infinity.. log[sin(1/n)]
A). Since none of the rules apply to this problem, plug it into your calculator in the table.
B). Your answer is 0.
When plugging into your calculator, look at the table and see what the number is approaching. For instance, if it’s .01, … , .001, …, .0001 then your answer would be 0.
Saturday, November 20, 2010
thirteen six.
chapter thirteen six.
this week we began chapter thirteen. this is the last chapter before our accumulated midterms. chapter thirteen explains sequences and series. there are many type of sequences but the most common are arithmetic (the adddition sequences) and geometric (the multiplication sequences).
section 13-6 is on the sigma. The sigma is apart of the Greek alphabet, but in math it is used to describe a condensed series. the sigma has three parts.. the top, bottom, and he right side. at the top of the sigma is the limit of summation. at the bottom is k= # where the "k" is the index and the numbers located at both the top and bottom of the sigma are limits of summation. to the side of the sigma is the summand,which is a function.
ex. expanding a sigma..
if k=1 ,the ending number ( above the sigma) is 3, and the summand is 2(k) then the expansion of the sigma would be :
2(1) + 2(2) + 2(3)+ 2(4) * simplify
2+4+6+8
this week we began chapter thirteen. this is the last chapter before our accumulated midterms. chapter thirteen explains sequences and series. there are many type of sequences but the most common are arithmetic (the adddition sequences) and geometric (the multiplication sequences).
section 13-6 is on the sigma. The sigma is apart of the Greek alphabet, but in math it is used to describe a condensed series. the sigma has three parts.. the top, bottom, and he right side. at the top of the sigma is the limit of summation. at the bottom is k= # where the "k" is the index and the numbers located at both the top and bottom of the sigma are limits of summation. to the side of the sigma is the summand,which is a function.
ex. expanding a sigma..
if k=1 ,the ending number ( above the sigma) is 3, and the summand is 2(k) then the expansion of the sigma would be :
2(1) + 2(2) + 2(3)+ 2(4) * simplify
2+4+6+8
Monday, November 15, 2010
Week 5 Blog Prompt
What is a famous sequences and series? What is it used for and who discovered it? Everyone should find a different type.
Sunday, November 14, 2010
13 - 1
This past week we started chapter thirteen. It is about sequences. You have to determine whether or not it is geometric, or arithmetic. There are formulas that you use to determine this. They are:
Arithmetic – tn = t1 + (n – 1)d
Geometric – tn = t1 x r^n-1
Example 1:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
17, 21, 25, 30. . .
It is neither because you do not subtract or add the same number every time. You cannot find a formula, because it is neither geometric or arithmetic.
Example 2:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
4, 6, 8, 10, 12. . .
This sequence is arithmetic because you add 2 every time. To find the formula, you plug into the arithmetic formula. d = 2, because that is the number you add everytime.
Tn = 4 + (n- 1)2
Tn= 4 + 2n -2
Tn =2 + 2n <--the formula you find.
Example 3:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
16, 8, 4, 2. . .
This sequence is geometric because if you put 2 over 4, 4 over 8, and 8 over 16, you get ½. r = ½, because that is the number you multiply by to get the sequence.
Tn = 16 x (1/2) ^n-1 <--the formula you find.
Arithmetic – tn = t1 + (n – 1)d
Geometric – tn = t1 x r^n-1
Example 1:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
17, 21, 25, 30. . .
It is neither because you do not subtract or add the same number every time. You cannot find a formula, because it is neither geometric or arithmetic.
Example 2:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
4, 6, 8, 10, 12. . .
This sequence is arithmetic because you add 2 every time. To find the formula, you plug into the arithmetic formula. d = 2, because that is the number you add everytime.
Tn = 4 + (n- 1)2
Tn= 4 + 2n -2
Tn =2 + 2n <--the formula you find.
Example 3:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
16, 8, 4, 2. . .
This sequence is geometric because if you put 2 over 4, 4 over 8, and 8 over 16, you get ½. r = ½, because that is the number you multiply by to get the sequence.
Tn = 16 x (1/2) ^n-1 <--the formula you find.
13-1
This week in advanced math we took a test on chapter 11 which involved lots of formulas. For the most part the test were easy as long as you knew the formulas. But at the end of the week we learned 13-1 which involves number sequences. There's arithmetic sequence, a sequence where the same number is added each time. And geometric, a sequence where the same number is multiplied each time.
Formulas:
arithmetic- tn=t1 + (n-1)d
geometric- tn= t1 * rn^-1
I don't really understand this yet so I don't know if this is right.
Ex.
How many multiples of 5 are there between 25 and 75
30, 35 , ... 75
d=9
75= 25 + (n-1)(9)
50=(n-1)(9)
49=(n)(9)
n=7
Playoffs Week 2
Well everyone, we are coming off a big win over Mangham. Now we play away at Pope John Paul in Slidell. Rebels all the way baby. Now, I don't remember much about this week. We did take a few tests. Not sure how good I did on those. Up now I remember. This week we learned about sequence. In the past sequences were always figured out with your brain and this is still the method I use, but there are formulsa which I was shocked to learn. This formulas completely simplify the process. I hope all of you had a good weekend. Peace.
13-1
This week we learned a section 1 in chapter 13. 13-1 was all about a list of numbers better known as sequence.
There are two sequences in 13-1:
Arithmetic-a sequence where the same number is added each time
Geometric-a sequence where the same number is multiplied each time
Arithmetic formula
tn=t1+(n-1)d
Examples:
5 7 9 11
is it arithmetic, geometric, or neither?
It is arithmetic because you add 2.
D=2
T1=4 t2=10 t75=?
T75=4(75-1)(6)
T75=448
Geometric formula
tn=t1=r^n-1
Examples:
6 12 24 48
is it arithmetic, geometric, or neither?
Geometric because you multiply 2
R=2
Find t10 if t1=4 t2=12 and the sequence is geometric.
Tn=4*(3) ^10-1
4*3^9=78,732
This section is easy if you know your formulas and stuff and now the sequence stuff section 13-1 shouldn’t be hard at all. Remember some problems in 13-1 would not be geometric and arithmetic because it can be neither. Know all the steps of how to identify and solve both methods. The methods are the arithmetic and geometric sequences.
There are two sequences in 13-1:
Arithmetic-a sequence where the same number is added each time
Geometric-a sequence where the same number is multiplied each time
Arithmetic formula
tn=t1+(n-1)d
Examples:
5 7 9 11
is it arithmetic, geometric, or neither?
It is arithmetic because you add 2.
D=2
T1=4 t2=10 t75=?
T75=4(75-1)(6)
T75=448
Geometric formula
tn=t1=r^n-1
Examples:
6 12 24 48
is it arithmetic, geometric, or neither?
Geometric because you multiply 2
R=2
Find t10 if t1=4 t2=12 and the sequence is geometric.
Tn=4*(3) ^10-1
4*3^9=78,732
This section is easy if you know your formulas and stuff and now the sequence stuff section 13-1 shouldn’t be hard at all. Remember some problems in 13-1 would not be geometric and arithmetic because it can be neither. Know all the steps of how to identify and solve both methods. The methods are the arithmetic and geometric sequences.
Chapter 13- Section 1
Chapter 13, Section 1 was taught on Thursday. I was not in class on Thursday due to the volleyball game so I am way behind and kind of lost, unfortunately. I knew if I missed the class I would be lost, but i really wanted to go and support the volleyball girls. I have done the homework, and followed as you did examples on Friday but I still dont feel comfortable with what I've learned. I am not confident enough to make my own examples for this chapter so I will do a few that I feel are simpler and easier for me to do. I will be sure to ask many questions on Monday, to make sure i get clarification on the things i feel lost about. Chapter 13-1 applies finding arithmetic and geometric sequences.
When doing these problems, you will first need to identify the sequence and its type.
Arithmetic- when the sequence of numbers has the same number added to them. Arithmetic Formula: tn= t1 + (n – 1) (d)
Geometric- when the sequence of numbers has the same number multiplied to them.Geometric Formula: tn= t1 x r^n -1
Examples:
1.State whether it is arithmetic or geometric. 1, 2, 4, 8, 16
It is geometric because 2 is being multiplied
2. How many multiples of 5 are there between 30 and 525?
The sequence is: 30, 35, 40, 45,…..525
The numbers are being added, so you use the arithmetic formula
525 = 30 + (n-1) (5)525 = 30 + 5n – 5 525 = 25 + 5n500 = 5nN = 100
When doing these problems, you will first need to identify the sequence and its type.
Arithmetic- when the sequence of numbers has the same number added to them. Arithmetic Formula: tn= t1 + (n – 1) (d)
Geometric- when the sequence of numbers has the same number multiplied to them.Geometric Formula: tn= t1 x r^n -1
Examples:
1.State whether it is arithmetic or geometric. 1, 2, 4, 8, 16
It is geometric because 2 is being multiplied
2. How many multiples of 5 are there between 30 and 525?
The sequence is: 30, 35, 40, 45,…..525
The numbers are being added, so you use the arithmetic formula
525 = 30 + (n-1) (5)525 = 30 + 5n – 5 525 = 25 + 5n500 = 5nN = 100
13-1 Sequences
This Thursday when everybody was at the volleyball game, Lauren, Kaitlyn, Ty, and I learned about sequences. It was a very small class which actually made it easier to learn.This section isn't hard just has a lot of steps which is hard when you have the attention span of a fly.
So, a sequence is just a list of numbers. There are two types of sequences: arithemetic which is a sequence where the same number is added or subtracted each time and geometric which is a sequence where the same number is multiplied or divided each time.
tn = whichever term
To find t n in arithemetic- tn = t1+ t(n-1)d and in geometric- tn = rn-1
These formulas are helpful when you're trying to figure out whether a problem is arithemetic, geometric, or neither.
For example
9/2, 3, 2, and 4/3 is that sequence arithemetic, geometric, or neither?
First you must divide each number with the second number over the first.
3/9/2 = 6/9 = 2/3
If all of the numbers equal the same thing it's geometric and the r = 2/3.
So, a sequence is just a list of numbers. There are two types of sequences: arithemetic which is a sequence where the same number is added or subtracted each time and geometric which is a sequence where the same number is multiplied or divided each time.
tn = whichever term
To find t n in arithemetic- tn = t1+ t(n-1)d and in geometric- tn = rn-1
These formulas are helpful when you're trying to figure out whether a problem is arithemetic, geometric, or neither.
For example
9/2, 3, 2, and 4/3 is that sequence arithemetic, geometric, or neither?
First you must divide each number with the second number over the first.
3/9/2 = 6/9 = 2/3
If all of the numbers equal the same thing it's geometric and the r = 2/3.
Chapter 13-1 was all about sequences-a list of numbers. There are 2 types of sequences: arithmetic and geometric.
ARITHMETIC:
a sequence where the same number is added each time.
FORMULA:
tn=t1+(n-1)d
EXAMPLES:
8, 10, 12, 14, 16
is it arithmetic, geometric, or neither?
its ARITHMETIC since 2 is being added each time.
d=2.
NOW FIND THE NTH TERM:
tn=8+(n-1)2
tn=8+2n-2
tn=2n+6.
GEOMETRIC:
a sequence where the same number is multiplied each time.
FORMULA:
tn=t1=r^n-1
EXAMPLES:
4, 8, 16, 32
is it arithmetic, geometric, or neither?
its GEOMETRIC because you can obviously tell that 2 is being multiplied.
so r=2.
I was at school thursday so I was able to learn this with you and it's actually really easy.
ARITHMETIC:
a sequence where the same number is added each time.
FORMULA:
tn=t1+(n-1)d
EXAMPLES:
8, 10, 12, 14, 16
is it arithmetic, geometric, or neither?
its ARITHMETIC since 2 is being added each time.
d=2.
NOW FIND THE NTH TERM:
tn=8+(n-1)2
tn=8+2n-2
tn=2n+6.
GEOMETRIC:
a sequence where the same number is multiplied each time.
FORMULA:
tn=t1=r^n-1
EXAMPLES:
4, 8, 16, 32
is it arithmetic, geometric, or neither?
its GEOMETRIC because you can obviously tell that 2 is being multiplied.
so r=2.
I was at school thursday so I was able to learn this with you and it's actually really easy.
To be honest I wish i knew what was going on in class. Therefore, im not really sure what to blog about. I would love to explain something we learned that i know best, but it just doesn't make any sense to me. So, in that case i will just explain 13-1 the best i can.
In Chapter 13 it is all about sequences. A sequence is simply a list of numbers (ex 4,8,12) There are two types of sequences:
Arithmetic: a sequence where the same number is added each time
Geometric: a sequence where the same number is multiplied each time
The formula for each are:
Arithmetic--> tn+t1+(n-1)d
Geometric-->tn=t1-r^n-1
To help you out "d" is the sequence, for example d of 4,6,8,10... is 2 because you add 2 each time. "n" is the last number given, for example 4,10...75 n would be 75. "r" is the ratio, for example ratio of (4,12) would be 3 because you are multiplying by 3.
Examples:
1. Find the indicated term--- t1=4 t2=10 t75=? If it is an arithmetic term
All to do is find d and plug into formula.
t75=4+(75-1)6
t75=448
2.Find t10 if t1=4 t2=12 and the sequence is geometric.
t10=4*3^10-1
t10=4*3^9
t10=78,732
There are many different problems to work with sequences, but that is mostly what you do.
In Chapter 13 it is all about sequences. A sequence is simply a list of numbers (ex 4,8,12) There are two types of sequences:
Arithmetic: a sequence where the same number is added each time
Geometric: a sequence where the same number is multiplied each time
The formula for each are:
Arithmetic--> tn+t1+(n-1)d
Geometric-->tn=t1-r^n-1
To help you out "d" is the sequence, for example d of 4,6,8,10... is 2 because you add 2 each time. "n" is the last number given, for example 4,10...75 n would be 75. "r" is the ratio, for example ratio of (4,12) would be 3 because you are multiplying by 3.
Examples:
1. Find the indicated term--- t1=4 t2=10 t75=? If it is an arithmetic term
All to do is find d and plug into formula.
t75=4+(75-1)6
t75=448
2.Find t10 if t1=4 t2=12 and the sequence is geometric.
t10=4*3^10-1
t10=4*3^9
t10=78,732
There are many different problems to work with sequences, but that is mostly what you do.
13-1
This week in Advanced Math we are learning chapter 13 section 1. This chapter is extremely easy for me I think because you are just finding the next numbers in a or telling how many are in a sequence. Somewhere in that range. There are many different types of problems that you may run into when doing this section but since we have such a good math teacher she has showed us every kind there is that we need to know, well that is for our homework at least.. no telling what we will see on the test. :) Anyways here are the formulas and following the formulas will be some examples which are really easy to catch on to.
Formulas:
arithemetic: tn=t1+(n-1)d
geometric: tn=t1xr^n-1
Examples:
1. Tell weather each set of sequences is geometric, arithmetic, or neither.
a. 3,6,9,12,15...
aritmetic d=3
b. 4,12,16, 20...
geometric r=4
c. 11, 15, 16, 19...
neither
Formulas:
arithemetic: tn=t1+(n-1)d
geometric: tn=t1xr^n-1
Examples:
1. Tell weather each set of sequences is geometric, arithmetic, or neither.
a. 3,6,9,12,15...
aritmetic d=3
b. 4,12,16, 20...
geometric r=4
c. 11, 15, 16, 19...
neither
13-1
This past week in advanced math we took a test on chapter 11 and started chapter 13. On Thursday they learned section 13-1, but I was not there because I attended the volleyball game. On Friday when I came back I was kind of lost. I got the notes I missed and tried to teach myself section 13-1. It was kind of hard to understand so I will have to spend extra time on this chapter. Section 13-1 is about arithmetic and geometric sequences. Like the previous chapters it seems we use formulas. I tried doing the homework we were given over the weekend. I am going to try and work some example problems that I did understand.
Formulas:
Arithmetic – tn = t1 = (n – 1) d
Geometric – tn = t1 = r n-1
Examples:
Tell whether each sequence is arithmetic, geometric, or neither.
4n + 3 is arithmetic
It’s arithmetic because if you plug in the 4 terms you get 7, 11, 15, 19 and d = 4.
8 – 5n is geometric
It’s geometric because if you plug in the 4 terms you get 3, -2, -7, -12 and d = -5.
A sequence is usually neither if you plug in your terms and get a set of random numbers.
Formulas:
Arithmetic – tn = t1 = (n – 1) d
Geometric – tn = t1 = r n-1
Examples:
Tell whether each sequence is arithmetic, geometric, or neither.
4n + 3 is arithmetic
It’s arithmetic because if you plug in the 4 terms you get 7, 11, 15, 19 and d = 4.
8 – 5n is geometric
It’s geometric because if you plug in the 4 terms you get 3, -2, -7, -12 and d = -5.
A sequence is usually neither if you plug in your terms and get a set of random numbers.
13-1
On Thursday, we started learning chapter 13-1. Unfortunately I wasn’t there for 7th hour that day so I had to try and catch up on this section Friday. It didn’t work out too well and I am not very sure what we are doing. I have been trying to teach myself, but I feel like I need a review in class on Monday. Since I am still kind of lost, I’ll show an example of some easy problems that I can do on my own.
Chapter 13-1 is about finding arithmetic and geometric sequences. The first thing that is usually asked is for you to identify.
Arithmetic- when the sequence of numbers has the same number added to them. The formula is:
tn= t1 + (n – 1) (d)
Geometric- when the sequence of numbers has the same number multiplied to them. The formula is:
tn= t1 x r^n -1
Examples:
1.State whether it is arithmetic or geometric.
1, 2, 4, 8, 16
Answer would be geometric because 2 is being multiplied.
2. How many multiples of 5 are there between 30 and 525?
The sequence is: 30, 35, 40, 45,…..525
The numbers are being added, so you use the arithmetic formula.
525 = 30 + (n-1) (5)
525 = 30 + 5n – 5
525 = 25 + 5n
500 = 5n
N = 100
Chapter 13-1 is about finding arithmetic and geometric sequences. The first thing that is usually asked is for you to identify.
Arithmetic- when the sequence of numbers has the same number added to them. The formula is:
tn= t1 + (n – 1) (d)
Geometric- when the sequence of numbers has the same number multiplied to them. The formula is:
tn= t1 x r^n -1
Examples:
1.State whether it is arithmetic or geometric.
1, 2, 4, 8, 16
Answer would be geometric because 2 is being multiplied.
2. How many multiples of 5 are there between 30 and 525?
The sequence is: 30, 35, 40, 45,…..525
The numbers are being added, so you use the arithmetic formula.
525 = 30 + (n-1) (5)
525 = 30 + 5n – 5
525 = 25 + 5n
500 = 5n
N = 100
13-1
We began Chapter 13 this week in Advanced Math. I3-1 is about arithmetic and geometric sequences. This section’s objective was to identify an arithmetic or geometric sequence and find a formula for its nth term.
Arithmetic Sequences - a sequence where the same number is added each time.
(Formula: tn = t1 + (n-1) d) The following sequences are all arithmetic..
Ex: 2, 6, 10, 14, 18.. – difference = 4
Geometric Sequences – a sequence where the same number is multiplied each time.
(Formula: tn = t1 x r^n-1)The following sequences are all geometric..
Ex: 1, 3, 9, 27, 81.. – ratio = 3
EX 1: 2, 5, 7, 10, 12.. Is it arithmetic, geometric, or neither?
A). It would be neither since you are adding 2 and 3 at different times. It isn’t a constant number being added or multiplied.
EX 2: How many multiples of 6 are there between 24and 300?
A). First find the multiples – 24, 36, 42, 48, 56.. 300
B). Since you are adding 6 each time, your difference = 4.
C). Plug the numbers into the arithmetic formula – tn = t1 + (n-1) d
D). 300 = 24 + (n-1)6
E). Solve using order of operations.
F). Your answer is n=47
EX3: Find t9 if t1=3, t2=6 and the sequence is geometric.
A). Since the problem is telling us it’s geometric, use that formula - tn = t1 x r^n-1
B). t9 = 3 x 2^9-1
C). t9 = 768
Arithmetic Sequences - a sequence where the same number is added each time.
(Formula: tn = t1 + (n-1) d) The following sequences are all arithmetic..
Ex: 2, 6, 10, 14, 18.. – difference = 4
Geometric Sequences – a sequence where the same number is multiplied each time.
(Formula: tn = t1 x r^n-1)The following sequences are all geometric..
Ex: 1, 3, 9, 27, 81.. – ratio = 3
EX 1: 2, 5, 7, 10, 12.. Is it arithmetic, geometric, or neither?
A). It would be neither since you are adding 2 and 3 at different times. It isn’t a constant number being added or multiplied.
EX 2: How many multiples of 6 are there between 24and 300?
A). First find the multiples – 24, 36, 42, 48, 56.. 300
B). Since you are adding 6 each time, your difference = 4.
C). Plug the numbers into the arithmetic formula – tn = t1 + (n-1) d
D). 300 = 24 + (n-1)6
E). Solve using order of operations.
F). Your answer is n=47
EX3: Find t9 if t1=3, t2=6 and the sequence is geometric.
A). Since the problem is telling us it’s geometric, use that formula - tn = t1 x r^n-1
B). t9 = 3 x 2^9-1
C). t9 = 768
Chapter 13-1
This week we began learning chapter thirteen, Sequences and Series. Due to certain circumstances, we only got to learn the first section “Finite sequences and series.” I found this section to be a little difficult. And the fact that the book doesn’t give the best examples did not help either. So I’m pretty stuck until Monday when we review. Until then, I guess ill keep doing the problems until I get it.
A sequence is a list of numbers. There are two kinds of sequences : Arithmetic and Geometric. In a arithmetic sequence, the same numbers are being added. On the other hand, in a geometric sequence, the same numbers are being multiplied. Both sequences have formulas that may be used to find nth terms.
Arithmetic- t(n)= t(1) + (n-1) d * here “d” is the common difference.
Geometric- t(n) = t(1)xR^(n-1) * here “r” is the common ratio.
* if a sequence is not arithmetic or geometric, then it is “Neither.”
There are many ways you may be asked to solve sequence problems. The first way you may be asked to solve a sequence problem is to solve for the nth root. Other ways include solving for the t(n)term where the n will be given, and you may even be asked how many terms are in a given sequence.
Ex. Find the formula for t(n)
1, 4, 7, 10..
Each number is being increased by three, therefore this sequence is an Arithmetic one. The common difference is a positive four. ( d=4 ) and it uses the arithmetic formula (t(n)= t(1) + (n-1) d ). To solve, you simply substitute. .
T(n)=1 + ( n-1)3
T(n)=1 + 3n-3
T(n) =3-2n
A sequence is a list of numbers. There are two kinds of sequences : Arithmetic and Geometric. In a arithmetic sequence, the same numbers are being added. On the other hand, in a geometric sequence, the same numbers are being multiplied. Both sequences have formulas that may be used to find nth terms.
Arithmetic- t(n)= t(1) + (n-1) d * here “d” is the common difference.
Geometric- t(n) = t(1)xR^(n-1) * here “r” is the common ratio.
* if a sequence is not arithmetic or geometric, then it is “Neither.”
There are many ways you may be asked to solve sequence problems. The first way you may be asked to solve a sequence problem is to solve for the nth root. Other ways include solving for the t(n)term where the n will be given, and you may even be asked how many terms are in a given sequence.
Ex. Find the formula for t(n)
1, 4, 7, 10..
Each number is being increased by three, therefore this sequence is an Arithmetic one. The common difference is a positive four. ( d=4 ) and it uses the arithmetic formula (t(n)= t(1) + (n-1) d ). To solve, you simply substitute. .
T(n)=1 + ( n-1)3
T(n)=1 + 3n-3
T(n) =3-2n
Thursday, November 11, 2010
Ch.11-1
The Chapter that i understood the most so far is Chapter 11 lesson 1. In Ch.11-1, we learned how to plot, give polar coordinates and convert to rectangular. I feel most comfortable converting to rectangular versus the others.
The Formula for converting to rectangular is x=rcos(theta) and y=rsin(theta).
To first convert to rectangular, you must plug in the numbers into the formula. You then solve for cos(ex:2COS90)after you solve cos( already knowing your trig chart) you multiply the number times the answer you get from the second step and you have half of the complete answer. you repeat the exact same steps for solving sin and you will have your complete answer.
EX: Convert(2,30 degrees) to rectangular
1. x=2cos30
2.cos30=square root of 3/2
3. 2 times square root of 3/2= square root of 3
4.(square root of 3,?)
SIN
1.y=2sin30
2.sin30=1/2
3.2 times 1/2=1
Answer:( square root of 3,1)
Week 4 Blog Prompt
What trig concept do you feel most comfortable with from Ch. 9-11? Give an example w/an explanation of how it is worked.
Sunday, November 7, 2010
Chapter 11-1
This past week in advance math, we finished learning chapter 11,consisting of formulas and trig chart memory.Two formulas we learned were x=rcostheta y=rsintheta(rectangular) r=sq/root x^2+y^2.tan theta=y/x(polar)
Ex:1: Give the polar coordinates for (2,4)
1. convert to polar using light green formula
2.r=square root of -1^2+12^2
3,r=+/-12
4.Plug x and y into the tan function Tantheta=4/-2
5/Tan i s negative, so now we are finding angles in quad 2 and 4..181 deg/ and 261 deg.
5.answer is ( 12,181 deg.) and (,-12-261)
Ex:2 convert (4,pi/2) to rect.
1.use formula in orange
2.x=4cos pi/2 and y=3 sin pi/2
3.use trig chart..pi/2 is (sin)sq.root of 2 over 2(cos)0
4.x=8sq.root of 2 y=0
5.(8sq.rtof 2,0)
Ex:1: Give the polar coordinates for (2,4)
1. convert to polar using light green formula
2.r=square root of -1^2+12^2
3,r=+/-12
4.Plug x and y into the tan function Tantheta=4/-2
5/Tan i s negative, so now we are finding angles in quad 2 and 4..181 deg/ and 261 deg.
5.answer is ( 12,181 deg.) and (,-12-261)
Ex:2 convert (4,pi/2) to rect.
1.use formula in orange
2.x=4cos pi/2 and y=3 sin pi/2
3.use trig chart..pi/2 is (sin)sq.root of 2 over 2(cos)0
4.x=8sq.root of 2 y=0
5.(8sq.rtof 2,0)
OMG! I UNDERSTAND!
So this weekend, I got some mathematical assistance from my twin, Mary Graci, the math whiz. She helped me better understand chapter 11. One of the things from chapter 11 is converting rectangular to polar and polar to rectangular.
The formula for converting to rectangular is
x = r cos theta
y = r sin theta
The formula for converting to polar is
r = square root of (x^2 + y^2)
theta = tan ^-1 (y/x)
The format for polar is
(r,theta)
The format for rectangular is
(x,y)
This may look confusing, but if you take it one step at a time it really isn't that scary.
For example, give the polar coordinates for (3,3).
First, you must find r.
r= square root of ( 3^2 + 3^2)
r = square root of 18
Now that you have r, you need to find theta.
theta= tan ^-1 (3/3) which is equal to one
And tan of one is 45 according to the trig chart.
So that makes your theta = 45
The answer you get is (square root of 18, 45 degrees) and (-square root of 18, 45 degrees).
Now, here is an example for converting to rectangular
Find (-2,60 degrees) in rectangular
x = -2 cos 60
y = -2 sin 60
x = -1
y = - square root of 3
Answer
(-1, -square root of 3)
The formula for converting to rectangular is
x = r cos theta
y = r sin theta
The formula for converting to polar is
r = square root of (x^2 + y^2)
theta = tan ^-1 (y/x)
The format for polar is
(r,theta)
The format for rectangular is
(x,y)
This may look confusing, but if you take it one step at a time it really isn't that scary.
For example, give the polar coordinates for (3,3).
First, you must find r.
r= square root of ( 3^2 + 3^2)
r = square root of 18
Now that you have r, you need to find theta.
theta= tan ^-1 (3/3) which is equal to one
And tan of one is 45 according to the trig chart.
So that makes your theta = 45
The answer you get is (square root of 18, 45 degrees) and (-square root of 18, 45 degrees).
Now, here is an example for converting to rectangular
Find (-2,60 degrees) in rectangular
x = -2 cos 60
y = -2 sin 60
x = -1
y = - square root of 3
Answer
(-1, -square root of 3)
11-1 Polars
This week in advanced math we learned chapter 11. We started by learning all of the different Polar graphs. There is Circles in Polar form where the formula is r= a cos(theta) or r= a sin(theta). Limacons where r=a +/- b cos(theta) or sin(theta) where a>0 and b>0. Cardiods where r=a +/- a sin(theta) or cos(theta). Rose Curves where r= a sin n(theta) or a cos n(theta). Lemniscates where r^2= a^2 sin 2(theta) or cos 2(theta). Then there's 11-2 with complex numbers. Which involves the formulas z= x + yi for rectangular and z= rcos(theta) + rsin(theta)i for Polar. The abbreviated version of this formula is z=rcis(theta). 11-3 involves De Moirre's Theorem with the formula (rcis)^n = r^n cisn(theta).
11-1 Polars
Formulas: x=rcos(theta); y=rsin(theta) to convert to rectangular
r=the square root of x^2 + y^2; tan(theta)=y/x to convert to polar
Ex. convert 2, 45 degrees to rectangular
x= 2cos45 degrees = 2(square root of 2)/2= square root of 2
y=2sin45degrees= 2(square root of 2)/2 = square root of 2
[square root of 2, square root of 2]
11-1
This week in advanced math we started and finished chapter 11. Chapter 11 consist of three sections that deal with polar, complex numbers and De Moivre’s Theorem. Section 11-1 was the easiest so let’s review. All you need to know is how to convert to rectangular and polar.
Formulas:
Convert to Rectangular
x = r cos θ
y = r sin θ
Convert to Polar
r = square root x2 + y2
tan θ = y/x
Let’s try some example problems
Example 1:
Convert (4, 45°) to rectangular
x = 4 cos 45° = 4(square root 2/2) = 4
y = 4 sin 45° = 4(square root 2/2) = 4
Your answer (4,4)
Example 2:
Convert (2, π/3) to rectangular
x = 2 cos 60° = 2 (1/2) = 1
y = 2 sin 60° = 2 (square root 2/2) = square root 3
Your answer (1, square root 3)
Example 3:
Give the polar coordinates (6, 8)
1. r = square root 62 + 82
2. square root 100 = +/- 10
3. tan θ = 8/6 reduces to 4/3
4. θ = tan-1(4/3) = 53.130°
5. You want 2 answers so add 180 and get 233.130°
Formulas:
Convert to Rectangular
x = r cos θ
y = r sin θ
Convert to Polar
r = square root x2 + y2
tan θ = y/x
Let’s try some example problems
Example 1:
Convert (4, 45°) to rectangular
x = 4 cos 45° = 4(square root 2/2) = 4
y = 4 sin 45° = 4(square root 2/2) = 4
Your answer (4,4)
Example 2:
Convert (2, π/3) to rectangular
x = 2 cos 60° = 2 (1/2) = 1
y = 2 sin 60° = 2 (square root 2/2) = square root 3
Your answer (1, square root 3)
Example 3:
Give the polar coordinates (6, 8)
1. r = square root 62 + 82
2. square root 100 = +/- 10
3. tan θ = 8/6 reduces to 4/3
4. θ = tan-1(4/3) = 53.130°
5. You want 2 answers so add 180 and get 233.130°
8th seed baby.
Thats right your Riverside Rebels are seeded 8th and that means we have a home playoff game this week. We are playing Mangum. Get ready people its going to be a fun ride. So on to Math. This week we started Chapter 11. This chapter comtains things called polar and rectangular and polar. Rectangular may be easily recognized on account of they look like the regular points on a coordinate plane (x,y). Polar is represented by a radius, which looks like a regular number, and and angle, which is normally represented by theta. We learned this week how to change rectangular to polar and vice versa. Another thing we did this week was learn how to classify graphs. Their can be circles, roses, limacon, if that politically correct, and some other stuff. Limacons are also called snails. I have to say that is pretty legit.
11-1
This week we are learning chapter 11. Chapter 11-1 is converting from polar to rectangular and converting from rectangular to polar. Also you will have to know how to plot the point in polar. I am going to show you how to do this, and give you some example problems and formulas.
FORMULAS:
(r, theta) is polar
(x,y) is rectangular
x=rcos(theta) and y=rsin(theta) is when you are converting to rectangular
r= +/-squareroot(x^2+y^2) and tan(theta) which is y/x is converting into polar
EXAMPLES:
Polar:
Give the polar point for (1,2) and plot the point in polar.
r=squareroot(1^2+2^2)
+/-5 which is going to be your first point, the x point
tan(theta)= 2/1
theta=tan^-1(2)
theta= 63.435
tan is postive in quadrants 1 and 3
Q1(5, 63.435)
Q3(5, 243.435)
To plot the polar point you just draw a number line go to 5(your x) and then do a unit circle of about how much you think 63.435 is and you do the same for the second quadrant.
Rectangular:
Give the rectangular coordinates for the point (4, 45degrees)
x=4cos45
y=4sin45
(2squareroot(2), 2squareroot(2))
FORMULAS:
(r, theta) is polar
(x,y) is rectangular
x=rcos(theta) and y=rsin(theta) is when you are converting to rectangular
r= +/-squareroot(x^2+y^2) and tan(theta) which is y/x is converting into polar
EXAMPLES:
Polar:
Give the polar point for (1,2) and plot the point in polar.
r=squareroot(1^2+2^2)
+/-5 which is going to be your first point, the x point
tan(theta)= 2/1
theta=tan^-1(2)
theta= 63.435
tan is postive in quadrants 1 and 3
Q1(5, 63.435)
Q3(5, 243.435)
To plot the polar point you just draw a number line go to 5(your x) and then do a unit circle of about how much you think 63.435 is and you do the same for the second quadrant.
Rectangular:
Give the rectangular coordinates for the point (4, 45degrees)
x=4cos45
y=4sin45
(2squareroot(2), 2squareroot(2))
11-1
This week we learned section 1 in chapter 11. 11-1 is really easy and quite simple. You will have to know formulas such as polar and rectangular.
To convert to rectangular you do x=rcostheta, y=rsintheta.
To convert to polar you do r=square roots of xsquared+ysquared and after you do tantheta=y/x then you find the inverse of it.
(r,theta)=polar, (x,y)=rectangular
Example Problems:
Give the Polar coordinates for the point (3,4)
R=square root 3squared+4squared=
R square root of 25 which is equal to +-5
Tantheta=4/3=
Theta=tan inverse of 4/3 you get 53.130degrees.
Where tan is + is 1st quadrant and 3rd quadrant which u do 180+53.130 =233.130degrees
Your answers are 5, 53.130degrees and -5,233.130degrees
Convert 2, 30 degrees to rectangular
X=2cos30 degrees which you get 2(square root of 3 over 2 because 30 degrees is on the trig chart you get square root of 3 as you answer.
Y=2sin30 degrees
Sin 30degrees is on the trig chart =to ½
2(1/2 you get 1
Your answers are squared root of 3 and 1.
If you know to convert to polar and rectangular and know every formula in 11-1 this section should be a joke for you.
To convert to rectangular you do x=rcostheta, y=rsintheta.
To convert to polar you do r=square roots of xsquared+ysquared and after you do tantheta=y/x then you find the inverse of it.
(r,theta)=polar, (x,y)=rectangular
Example Problems:
Give the Polar coordinates for the point (3,4)
R=square root 3squared+4squared=
R square root of 25 which is equal to +-5
Tantheta=4/3=
Theta=tan inverse of 4/3 you get 53.130degrees.
Where tan is + is 1st quadrant and 3rd quadrant which u do 180+53.130 =233.130degrees
Your answers are 5, 53.130degrees and -5,233.130degrees
Convert 2, 30 degrees to rectangular
X=2cos30 degrees which you get 2(square root of 3 over 2 because 30 degrees is on the trig chart you get square root of 3 as you answer.
Y=2sin30 degrees
Sin 30degrees is on the trig chart =to ½
2(1/2 you get 1
Your answers are squared root of 3 and 1.
If you know to convert to polar and rectangular and know every formula in 11-1 this section should be a joke for you.
Chapter Eleven
This week our class began chapter eleven and learned chapter complex numbers. Complex numbers have two forms : polar and rectangular.
The rectangular form for complex numbers is z= x + yi
The polar form for complex numbers is z= rcos(x)rsin(x)i or it can be abbreviated as z=rcis(x)
In section one we learned how to convert polar to non polar and vice versa.
Rectangular coordinates are formed by ( x,y)
Polar coordinates are in the form of (r,θ)
In order to convert from rectangular to polar you must solve for “r” and theta.
*r=sqrt of x^2+y^2. Theta is equal to tan y/x
You will always have two answers when converting to polar. One positive and one negative.
EX. A (3,3) CONVERT TO POLAR :
R=sqrt 3^2=3^2 = 9+9
R=sqrt18 = ±2√3
Theta= tan y/x
theta= tan(3/3) = tan 1
theta = 45, 225
*(3,3) is located in the first two quadrants therefore the positive number gets the smaller number. Your answers are (2, 45) (-2,225)
In order to convert to rectangular , you simply use formulas x= rcostheta and y=rsintheta
The rectangular form for complex numbers is z= x + yi
The polar form for complex numbers is z= rcos(x)rsin(x)i or it can be abbreviated as z=rcis(x)
In section one we learned how to convert polar to non polar and vice versa.
Rectangular coordinates are formed by ( x,y)
Polar coordinates are in the form of (r,θ)
In order to convert from rectangular to polar you must solve for “r” and theta.
*r=sqrt of x^2+y^2. Theta is equal to tan y/x
You will always have two answers when converting to polar. One positive and one negative.
EX. A (3,3) CONVERT TO POLAR :
R=sqrt 3^2=3^2 = 9+9
R=sqrt18 = ±2√3
Theta= tan y/x
theta= tan(3/3) = tan 1
theta = 45, 225
*(3,3) is located in the first two quadrants therefore the positive number gets the smaller number. Your answers are (2, 45) (-2,225)
In order to convert to rectangular , you simply use formulas x= rcostheta and y=rsintheta
11-1
This week we started learning chapter 11. A lot of people thought this chapter was sort of easy, but I didn’t. I think this is one of the hardest chapters we have done so far, and doing the pages in the book for a grade really stresses me out because they are a lot harder and will probably bring my grade down. But, I did understand section 11-1 the best out of all of them. For this, you need to be able to convert things into either polar or rectangular.
Convert to rectangular – x = r cos theta and y = r sin theta
Convert to polar- square root x^2 = y^2 and tan theta = y/x
For these problems, you will be given points.
Example 1.
Give polar coordinates for (1, 1)
1. R= square root 1^2 + 1^2
2. R = +/- square root 5
3. Tan theta = 1/1
4. Theta equals tan^-1 (1/1) = 1
5. Since tan is positive, it goes in the first and third quadrants
6. Quadrant 1 = 45 (from the trig chart) quadrant 3 = 225
7. Answer= (square root 2, 45) and ( -square root 2, 225)
Example 2.
Give rectangular coordinates for (1, pi/6)
1. X = 1cos pi/6 and y= 1sin pi/6
2. These are in your trig chart, 1(0) and 1(1)
3. The answer is (0, 1)
Convert to rectangular – x = r cos theta and y = r sin theta
Convert to polar- square root x^2 = y^2 and tan theta = y/x
For these problems, you will be given points.
Example 1.
Give polar coordinates for (1, 1)
1. R= square root 1^2 + 1^2
2. R = +/- square root 5
3. Tan theta = 1/1
4. Theta equals tan^-1 (1/1) = 1
5. Since tan is positive, it goes in the first and third quadrants
6. Quadrant 1 = 45 (from the trig chart) quadrant 3 = 225
7. Answer= (square root 2, 45) and ( -square root 2, 225)
Example 2.
Give rectangular coordinates for (1, pi/6)
1. X = 1cos pi/6 and y= 1sin pi/6
2. These are in your trig chart, 1(0) and 1(1)
3. The answer is (0, 1)
Chapter 11- Section 1
This week in Advanced Math, we were introduced to Chapter 11. To be honest i feel this chapter is quite simple. Calculators are still forbidden of use on test but we were allowed to use them for homework. These chapters all feed off of one another, and are accumlative. In 11-1, we were taught two conversion formulas – polar and rectangular.
x= r cos theta and y = r sin theta (formulas used to convert to Rectangular form)
r = square root x^2 + y^2 and tan theta = y/x (formulas used to convert to Polar form)
Points will be given to you like this:
(x , y ) <----Rectangular
(r, theta) <------Polar
EXAMPLE 1: Give the polar coordinates for point (-3, 4).
1. Since we are converting to polar, use the pink formulas.
2. r = square root of -3^2 + 4^2
3. r = +/- 5
4. (Plug x and y into the tan function y/x) tan theta = 4/-3
5. Since tan is negative, find the angles in quadrants 2 and 4 – 126.87 degrees and 306.87 degrees
6. Since the angles are negative, +5 will get the bigger angle.
7. Your answer is (5, 306.87 degrees) and (-5, 126.87 degrees)
EXAMPLE 2: Convert (3, pi/2) to rectangular.
1. Since we are converting to rectangular, use the green formulas.
2. x = 3 cos pi/2 and y = 3 sin pi/2
3. Pi/2 is in your trig chart so continue simplifying.
4. x = 3(0) and y = 3(1)
5. Your answer is (0,3)
x= r cos theta and y = r sin theta (formulas used to convert to Rectangular form)
r = square root x^2 + y^2 and tan theta = y/x (formulas used to convert to Polar form)
Points will be given to you like this:
(x , y ) <----Rectangular
(r, theta) <------Polar
EXAMPLE 1: Give the polar coordinates for point (-3, 4).
1. Since we are converting to polar, use the pink formulas.
2. r = square root of -3^2 + 4^2
3. r = +/- 5
4. (Plug x and y into the tan function y/x) tan theta = 4/-3
5. Since tan is negative, find the angles in quadrants 2 and 4 – 126.87 degrees and 306.87 degrees
6. Since the angles are negative, +5 will get the bigger angle.
7. Your answer is (5, 306.87 degrees) and (-5, 126.87 degrees)
EXAMPLE 2: Convert (3, pi/2) to rectangular.
1. Since we are converting to rectangular, use the green formulas.
2. x = 3 cos pi/2 and y = 3 sin pi/2
3. Pi/2 is in your trig chart so continue simplifying.
4. x = 3(0) and y = 3(1)
5. Your answer is (0,3)
Saturday, November 6, 2010
11-1
This week in Advanced Math, we covered all of Chapter 11. The whole Chapter consisted of formulas and was pretty simple. In 11-1, we were taught two conversion formulas – polar and rectangular.
x= r cos theta
y = r sin theta
(r, theta)
^ Used to convert to rectangular.
r = square root x^2 + y^2
tan theta = y/x
(x,y)
^Used to convert to polar
EXAMPLE 1: Give the polar coordinates for point (-3, 4).
1. Since we are converting to polar, use the red formulas.
2. r = square root of -3^2 + 4^2
3. r = +/- 5
4. Plug x and y into the tan function – tan theta = 4/-3
5. Since tan is negative, find the angles in quadrants 2 and 4 – 126.87 degrees and 306.87 degrees
6. Since the angles are negative, +5 will get the bigger angle.
7. Your answer is (5, 306.87 degrees) and (-5, 126.87 degrees)
EXAMPLE 2: Convert (3, pi/2) to rectangular.
1. Since we are converting to rectangular, use the blue formulas.
2. x = 3 cos pi/2 and y = 3 sin pi/2
3. Pi/2 is in your trig chart so continue simplifying.
4. x = 3(0) and y = 3(1)
5. Your answer is (0,3)
x= r cos theta
y = r sin theta
(r, theta)
^ Used to convert to rectangular.
r = square root x^2 + y^2
tan theta = y/x
(x,y)
^Used to convert to polar
EXAMPLE 1: Give the polar coordinates for point (-3, 4).
1. Since we are converting to polar, use the red formulas.
2. r = square root of -3^2 + 4^2
3. r = +/- 5
4. Plug x and y into the tan function – tan theta = 4/-3
5. Since tan is negative, find the angles in quadrants 2 and 4 – 126.87 degrees and 306.87 degrees
6. Since the angles are negative, +5 will get the bigger angle.
7. Your answer is (5, 306.87 degrees) and (-5, 126.87 degrees)
EXAMPLE 2: Convert (3, pi/2) to rectangular.
1. Since we are converting to rectangular, use the blue formulas.
2. x = 3 cos pi/2 and y = 3 sin pi/2
3. Pi/2 is in your trig chart so continue simplifying.
4. x = 3(0) and y = 3(1)
5. Your answer is (0,3)
Tuesday, November 2, 2010
Week 3 Prompt
What are the different types of polar graphs? Give several examples with pictures or a site to use as a reference. Everyone should have different pictures or a different site.
Sunday, October 31, 2010
10-3 Simplify
This week in advanced math we learned chapter 10-3 and 10-4 along with reviewing chapter 10 and testing on it. In 10-1 you have the formulas sin(alpha + beta)= sin(alpha)cos(beta) + cos(alpha)sin(beta) and cos(alpha + beta)= cos(alpha)cos(beta) - sin(alpha)sin(beta). In 10-2 you have the tangent formula tan(alpha + beta)= tan(alpha) + tan(beta)/1-tan(alpha)tan(beta). Then there's 10-3.
10-3 Formulas
Doouble Formulas
Sin 2(alpha)= 2sin(alpha)cos(alpha)
Cos2(alpha)= cos^2(alpha)-sin^2(alpha) or 1-2sin^2(alpha) or 2cos^2(alpha)-1
tan2(alpha)= 2tan(alpha)/1-tan^2(alpha)
Half Formulas
sin alpha/2=+/- the square root of 1-cos(alpha)/2
cos alpha/2= +/- the square root of 1+ cos(alpha)/2
tan alpha/2 = +/- the square root of 1-cos(alpha)/1+cos(alpha) or sin(alpha)/1+cos(alpha) or
1-cos(alpha)/sin(alpha)
Ex. Simplify 2sin(90 degrees)cos(90 degrees)
1) Sin2(90 degrees)
2)sin 180 degrees
3= 180-180= sin(0)
4)=0
happy boo day!
We did not learn to much new material this week. We had a chapter test this week on chapter ten, and had a sub for two days, the wonderful Ms. FeFe. I missed two days this week, which sort of put me behind, and for this reason i feel i didnt do to well with this chapter. We could ask questions on problems we were not sure how to do. 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
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
So this past week we didn't learn anything new. We basically just reviewed chapter 10 and took a chapter test. I tried my best to catch on to this chapter but i felt horrible all week. So here's some examples.
EXAMPLE 1:
sin75 degrees cos 15 degrees+cos 75 degrees sin 15 degrees
first you must replace it with a formula!
sin(alpha+beta)
sin (75+15)
=sin 90 degrees
TRIG CHART!
your answer is 1.
EXAMPLE 2:
cos 5pi/12 cos pi/12-sin 5pi/12 sin pi/12
replace formula!
cos(alpha+beta)
cos(5pi/12+pi/12)
cos 6pi/12
=1/2
TRIG CHART!
your answer is pi/3.
EXAMPLE 1:
sin75 degrees cos 15 degrees+cos 75 degrees sin 15 degrees
first you must replace it with a formula!
sin(alpha+beta)
sin (75+15)
=sin 90 degrees
TRIG CHART!
your answer is 1.
EXAMPLE 2:
cos 5pi/12 cos pi/12-sin 5pi/12 sin pi/12
replace formula!
cos(alpha+beta)
cos(5pi/12+pi/12)
cos 6pi/12
=1/2
TRIG CHART!
your answer is pi/3.
HAPPY HALLOWEEN!?!?!
First off I want to wish everyone a HaPpY hAlLoWeEn!!!!! This week in advanced math we did not learn anything new. We reviewed chapter 10 on Monday and Tuesday, took a chapter test on Wednesday and Thursday and had a sub Thursday and Friday. Our chapter test was on sections 1-4. The easiest section was 10.1 because you just have to know two formulas and you just have to plug it in. You also needed to know you trig chart.
The formulas for section 10-1 are:
Sin (α +/- β) = (sin α) (cos β) +/- (cos α) (sin β)
Cos (α +/- β) = (cos α) (cos β) -/+ (sin α) (sin β)
Find the exact value of sin 15degrees
Alpha=45 degrees
Beta= 30 degrees
Sin (45degrees-30degrees) = (sin 45degrees) (cos 30degrees)-(cos45degrees) (sin 30 degrees)
(Square root of 2 over 2)(Square root of 3 over 2) – (Square root of 2 over 2) (1/2)
Square root of 6 over 4 –square root of 2 over 4= square root of 6- square root of 2 over 4.
Your answer is Sin 15degrees= square root of 6-square root of 2 over 4.
Because it isn’t on trig chart u leave it at that if it was then you can simplify it.
You see how simple this section is.
Remember know your formulas and know your trig chart. You basically need to master your trig chart and know it in order to do some problems because some stuff are plug from trig chart like sin 45 for example is square root of 2 over 2. NO calculators !!!
The formulas for section 10-1 are:
Sin (α +/- β) = (sin α) (cos β) +/- (cos α) (sin β)
Cos (α +/- β) = (cos α) (cos β) -/+ (sin α) (sin β)
Find the exact value of sin 15degrees
Alpha=45 degrees
Beta= 30 degrees
Sin (45degrees-30degrees) = (sin 45degrees) (cos 30degrees)-(cos45degrees) (sin 30 degrees)
(Square root of 2 over 2)(Square root of 3 over 2) – (Square root of 2 over 2) (1/2)
Square root of 6 over 4 –square root of 2 over 4= square root of 6- square root of 2 over 4.
Your answer is Sin 15degrees= square root of 6-square root of 2 over 4.
Because it isn’t on trig chart u leave it at that if it was then you can simplify it.
You see how simple this section is.
Remember know your formulas and know your trig chart. You basically need to master your trig chart and know it in order to do some problems because some stuff are plug from trig chart like sin 45 for example is square root of 2 over 2. NO calculators !!!
Happy Halloween!
First off I want to wish everyone a HAPPY HALLOWEEN!!!!! This week in advanced math we did not learn anything new. We reviewed chapter 10 on Monday and Tuesday, took a chapter test on Wednesday and Thursday and had a sub Thursday and Friday. Our chapter test was on sections 1-4. All for sections had to do with memorizing formulas. If you did not use a formula for every problem on the test you did it wrong. Section 1 was the easiest section of the 4 because you just had to memorize 2formulas. Here are the sine and cosine formulas:
Sin (α +/- β) = (sin α) (cos β) +/- (cos α) (sin β)
Cos (α +/- β) = (cos α) (cos β) -/+ (sin α) (sin β)
Let’s try an example:
Find the exact value of sin 75°
1. Sin (45° + 30°) = (sin 45°) (cos 30°) + (cos 45°) (sin 30°)
2. (square root 2/2) (square root3/2) + (square root 2/2) (½)
3. (square root 6/4) + (square root 2/4)
4. square root 6 + square root 2/ 4
Sin (α +/- β) = (sin α) (cos β) +/- (cos α) (sin β)
Cos (α +/- β) = (cos α) (cos β) -/+ (sin α) (sin β)
Let’s try an example:
Find the exact value of sin 75°
1. Sin (45° + 30°) = (sin 45°) (cos 30°) + (cos 45°) (sin 30°)
2. (square root 2/2) (square root3/2) + (square root 2/2) (½)
3. (square root 6/4) + (square root 2/4)
4. square root 6 + square root 2/ 4
HAPPY HALLOWEEN!!!
So the week before last, I was crazy sick. This week people took tests and hung out with Mrs. FeFe. That means that I really didn't learn anything. I stole notes from Lauren Louque, but I really don't understand them that well.
Here's an example of things I have in my notes.
From 10-4
cos2x = cos x
2cos^2x-1 = cosx
2cos^2x-cosx-1 = 0
2cos^2x-2cosx+cosx-1
2cosx(cosx-1) + (cos-1)
(2cosx+1) (cosx-1)
cosx = 1/2
I'm still having trouble with this stuff. But hopefully I can get some help.
Anyway,
HAPPY HALLOWEEN!!!
Here's an example of things I have in my notes.
From 10-4
cos2x = cos x
2cos^2x-1 = cosx
2cos^2x-cosx-1 = 0
2cos^2x-2cosx+cosx-1
2cosx(cosx-1) + (cos-1)
(2cosx+1) (cosx-1)
cosx = 1/2
I'm still having trouble with this stuff. But hopefully I can get some help.
Anyway,
HAPPY HALLOWEEN!!!
Chapter 10 Section 1
Just as everybody else has been saying, we haven't really learned much this week, we just reviewed and tested and Chilled with FE FE. But one of the things that I ACTUALLY learned, was Section 10-1. Even though this sections does not permit the use of calculators, it still is very simple and easy to understand.
The formulas in 10-1 are as followed:
cos (alpha +/- beta) = cos alpha cos beta+/- sin alpha sin beta
sin (alpha +/- beta) = sin alpha sin beta +/- cos alpha cos beta
Now lets get to work!
Find the exact value of sin 15*
Alpha = 45 *
Beta = 30 *
Sin (45 * - 30 *) = (sin 45*) (cos 30*) - (cos 45*) (sin 30 *)
(Square root of 2 \ 2)(Square root of 3 \ 2) – (Square root of 2 \ 2) (1 \ 2)
Square root of 6 \ 4 – square root of 2 \ 4= square root of 6 - square root of 2 \ 4.
= Sin 15* = square root of 6-square root of 2 \ 4.
Alpha = 45 *
Beta = 30 *
Sin (45 * - 30 *) = (sin 45*) (cos 30*) - (cos 45*) (sin 30 *)
(Square root of 2 \ 2)(Square root of 3 \ 2) – (Square root of 2 \ 2) (1 \ 2)
Square root of 6 \ 4 – square root of 2 \ 4= square root of 6 - square root of 2 \ 4.
= Sin 15* = square root of 6-square root of 2 \ 4.
Now how easy was DAT! hehe
Also, here a little tip. Make sure you keep a keen eye for trig chart functions, if you miss one, then you may not be allowed to answer your problem correctly.
Happy Halloween
HAPPY HALLOWEEN EVERYONE :D
This week we didn’t really learn anything new. We reviewed and took our Chapter tests on Chapter 10. In 10-3 we were given double and half angle formulas. I would type all the formulas out, but I honestly don’t feel like it. Everyone should have them in their binders or even better, KNOW THEM by now.
EXAMPLE 1: 2sin30 degrees cos30 degrees
A). According to our formulas, this is the right side of the sin 2 alpha formula.
B). So, replace it with the left – sin 2(30 degrees)
C). Your answer is sin 60 degrees which is in your trig chart.
D). It reduces to square root of 3 / 2.
EXAMPLE 2: cos squared 15 degrees – sin squared 15 degrees
A). According to our formulas, this is the right side of the cos 2 alpha formula.
B). So, replace it with the left – cos 2(15 degrees)
C). Your answer is cos 30 degrees which is in your trig chart.
D). It reduces to squre root of 3 / 2.
Have a great weekenddddd!
This week we didn’t really learn anything new. We reviewed and took our Chapter tests on Chapter 10. In 10-3 we were given double and half angle formulas. I would type all the formulas out, but I honestly don’t feel like it. Everyone should have them in their binders or even better, KNOW THEM by now.
EXAMPLE 1: 2sin30 degrees cos30 degrees
A). According to our formulas, this is the right side of the sin 2 alpha formula.
B). So, replace it with the left – sin 2(30 degrees)
C). Your answer is sin 60 degrees which is in your trig chart.
D). It reduces to square root of 3 / 2.
EXAMPLE 2: cos squared 15 degrees – sin squared 15 degrees
A). According to our formulas, this is the right side of the cos 2 alpha formula.
B). So, replace it with the left – cos 2(15 degrees)
C). Your answer is cos 30 degrees which is in your trig chart.
D). It reduces to squre root of 3 / 2.
Have a great weekenddddd!
Subscribe to:
Posts (Atom)