Hi, Pier and I made our scribe in Acrobat, so, here's the link:
https://createpdf.adobe.com/cgi-pickup.pl/Scribe%20Post%20May%2031.pdf?BP=IE&LOC=en_US&CUS=c82e73ba86bf33f7252d632356571913&CDS=465F81FE-0261-082BBE
The homework on the board was: PG.500-503, PG.502, #1-19 ODD
We are in computer class, and I just relaized that you need the username and password to access it, while I can igure out how to copy the images from adobe into a normal post, you can use my Adobe Id: cristybustillo731@gmail.com, and password: gmbateri.
Sorry!!!
Thursday, May 31, 2007
MathCentre Web Page
Hi people!
I was searching in the internet for Finite Series and I found this great Math page where I found good information about the Sum of Infinite Series. It is really helpful and, in advantage, it is a great Math webpage where you can search more about any subject in Mathematics. I hope you fin it helpful.
Bee
http://www.mathcentre.ac.uk/students.php/all_subjects/series/convergence/resources/367
I was searching in the internet for Finite Series and I found this great Math page where I found good information about the Sum of Infinite Series. It is really helpful and, in advantage, it is a great Math webpage where you can search more about any subject in Mathematics. I hope you fin it helpful.
Bee
http://www.mathcentre.ac.uk/students.php/all_subjects/series/convergence/resources/367
Limit Solutions
Below you will find the solutions to the limit problems listed in Frank and Nico's post from a couple of days ago. Try solving them yourselves first.
Wednesday, May 30, 2007
*ScRiBe* May 30th, 2007 Gabriela Guerrero
Mpooh's POD:
- Write a formula for the nth term of the given infinite sequence:
e, -e4, e9, -e16,...
- Sketch a rough graph of the sequence.
- Take the problem steps by steps. Analysis the sequence and find what you know and can solve. For example, the exponents on e, the exponent is 12. 22, 32, 42... By knowing this we know that e^n2, gives the answers. So next to do if find out how to create the alteration of positive and negavite sings. (-1)n+1.
The answer Tn=(-1)n+1e^n2.
lets check:
T1=(-1)1+1(e1^2)
T1=(-1)2 (e1)
T1=e
Lim= DNE (each time the points on the graph are further apart.)
Sybil the Rat analysis:
S is a geometric series with r=1/2
Sn= distance covered= (1/2) 1-(1/2)^n/ 1-1/2
Since it is geometric we take the Lim of each side. Remember lim must be 1 in order for the rat to cross the room. (as n approches infinity.)
LimSn= Lim(1/2) 1-(1/2)^n/ 1-1/2
LimSn= [1-(1/2)^n]
LimSn= Lim(1-0)
LimSn=1
Tortoise and achilles Geometric series.
Asume that the tortoise gets a 10-meter head start and that Achilles runs 10 times as fast. At what distance and at what time will achilles catch the tortoise?
Tn= distance that achilles hast to run.)
first few terms: 10+ 1 + 1/10+ 1/100..+Sn..
r=1/10; T1=10
Sn=T1[1-(1/10)^n/(1-1/2)]
take the lim of each side, asume that n is approching infinity.
LimSn=lim[1-(1/10)^n/(1-1/10)]
¨(1/10)^n¨- based on the book theorem: If /r/less than 1, the Limr^n=o. as n approches infinity.
LimSn=lim[1/(1-1/10)]
LimSn= (10/9)/10
LimSn=100/9 = 11.1111 meters achilles will catch the tortoise.
Notice that the n in both problem the formula simplifies to:
Sn=T1/1-r
Theorem: Sum of an infinite geometric series:
If /r/ less than infinite geometric series converges to the sum.
Sn=T1/1-r ; T1+T2+T3+T4+...Tn
If /r/greater or equal to 1, and T1 is not equal to 0, then the series diverges. meaning that every time the points are getting further apart.
(Tortoise and achilles problem can be done with 2 linear equations.)
Mpooh Has Got Problems
Pooh says:
Hello, Since I am having some difficulties with Geometric Series problem, I thought It was kind of nice to post some problems, some which are very challenging, to those interested. The answers I did not post, but if anyone wants them, they can try the questions and are invited to ask me for the answers. Feel free and try these problem, post comments with your answers!!
Ready?
There are Sequence and Series problems! GOOD LUCK!
Find all of the terms of the finite sequence?
1a.
; 1 < n < 5
Find the first five terms and the twelfth term of the infinite sequence.
2a.
Write a formula for the nth term of given the infinite sequence.
3a.
1. Write a formula for the nth term of the geometric sequence 7, 28, 112, 448, .... Do not use a recursive formula.
2. Write a formula for the nth term of the geometric sequence 16, - 4, 1, -1/4, .... Do not use a recursive formula.
3. Find the first term of a geometric sequence with a fifth term of 32 and a common ratio of -2.
4. Find the common ratio for a geometric sequence with a first term of 3/4 and a third term of 27/16.
5. Find the sum of the finite geometric series 3 - 6 + 12 - 24 + 48 - 96.
6. Write a formula for the nth term of the given geometric sequence. Do not use a recursive formula.
125, 25, 5, 1, ...
4, -12, 36, -108, ...
7. Find the sum of the given finite geometric series.
2 + 14 + 98 + 686 + 4802 + 33614 + 235298
Hello, Since I am having some difficulties with Geometric Series problem, I thought It was kind of nice to post some problems, some which are very challenging, to those interested. The answers I did not post, but if anyone wants them, they can try the questions and are invited to ask me for the answers. Feel free and try these problem, post comments with your answers!!
Ready?
There are Sequence and Series problems! GOOD LUCK!
Find all of the terms of the finite sequence?
1a.
; 1 < n < 5
Find the first five terms and the twelfth term of the infinite sequence.
2a.
Write a formula for the nth term of given the infinite sequence.
3a.
1. Write a formula for the nth term of the geometric sequence 7, 28, 112, 448, .... Do not use a recursive formula.
2. Write a formula for the nth term of the geometric sequence 16, - 4, 1, -1/4, .... Do not use a recursive formula.
3. Find the first term of a geometric sequence with a fifth term of 32 and a common ratio of -2.
4. Find the common ratio for a geometric sequence with a first term of 3/4 and a third term of 27/16.
5. Find the sum of the finite geometric series 3 - 6 + 12 - 24 + 48 - 96.
6. Write a formula for the nth term of the given geometric sequence. Do not use a recursive formula.
125, 25, 5, 1, ...
4, -12, 36, -108, ...
7. Find the sum of the given finite geometric series.
2 + 14 + 98 + 686 + 4802 + 33614 + 235298
Tuesday, May 29, 2007
Daniel Enrique Rubio Scribe Post
Hi, It’s me, Daniel Rubio and it is my turn to be the scribe. Today’s POD was this one:
“Pier becomes possesed by the devil so that he can spin his head all the way around. Suppose that Pier’s head is a perfect sphere and that he sins his head at a constant rate of 8 revolutions per minute. A small ant is perched on Pier’s nose and in 6 seconds it travels 60 cm. Fin the radius of Pier’s head.”
A small hint: Find the centra angle that the ant sweeps.
As a first step lets make a picture to make understanding easier.
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Now that we have made ourselves clear lets look for ways to get to the solution of the problem. In class, we saw 3 different approaches to the problem: Susy DC’s, MAC’s , and Mr. Alcantara’s approach to the problem. Before startig to explain each way to solve the problem, it is important to know which data we have.
-Pier’s head is spinning at a constant rate of 8 revs/min
-The Fly traveled 60 cm in 6 sec ::: Important note: we can use this as sector length to solve the problem :::
Now that we’ve made that clear, we can start on the solutions.
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Now, moving forward we look into MAC’s approach; which has a similar reasoning but a different execution
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Now moving on to Mr Alcantara’s approach, which uses a simpler reasoning on the problem
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Now we are FINALLY done with the POD… it took me a while to do those ones.
I am now going to post about today’s ACTUAL lesson.
Sybil the rat : A mathematicla answer .
-Today we reviewed the problem of Sybil the rat : A mathematicla answer .
Sybil must 1st cross half of a room, the ¼ of it, then 1/8 and so on.
First thing we need to realize is, tn = fraction of room to cross. So when we are in doubt and don’t know what to do, we list the first terms of the sequence, which in this case are:
½ , ¼ , 1/8 , 1 /16 … as we could realize the explicit definition for this sequence is: n
(1/2)
>So we need to ask ourselves a question… what happens to tn as n approaches infinity?
First, we should give a graphical answer- REMEMBER that your grapgh shoud include
>A scale with labels
> tn values as it approaches its limit ( in this case zero)
> tn values as n approaches infinity
>must label both axes
>must show points not lines( remember it is a sequence)
>Interpret the graph
>Now what if we wanted to know the total amount of room crossed?
Sn= total amount of room crossed
If Sybil can cross the room then Sn> or equal to 1
List find first terms of the series.
Realize that it is a geometric series whichs r = 1/2
Well that was basically the most important part, Mr. Alcantara said something about using algebra on limits to cancel them out and get somenthing which I didn’t really understand. Mr. Alcantara said he did’t expect us to know it, so I’m not covering it now.
Homework was : Read 500-503
Pg. 502 1-11 odd
n
In the book they mention a new theorem, that if l r l < 1, then lim r = 0.
n->infinity
If you think about it it is totally logical because you are repeatedly multiplying by a number less than one, each time oits going to go closer to zero.
Well that’s it for today.
“Pier becomes possesed by the devil so that he can spin his head all the way around. Suppose that Pier’s head is a perfect sphere and that he sins his head at a constant rate of 8 revolutions per minute. A small ant is perched on Pier’s nose and in 6 seconds it travels 60 cm. Fin the radius of Pier’s head.”
A small hint: Find the centra angle that the ant sweeps.
As a first step lets make a picture to make understanding easier.
This album is powered by BubbleShare - Add to my blog
Now that we have made ourselves clear lets look for ways to get to the solution of the problem. In class, we saw 3 different approaches to the problem: Susy DC’s, MAC’s , and Mr. Alcantara’s approach to the problem. Before startig to explain each way to solve the problem, it is important to know which data we have.
-Pier’s head is spinning at a constant rate of 8 revs/min
-The Fly traveled 60 cm in 6 sec ::: Important note: we can use this as sector length to solve the problem :::
Now that we’ve made that clear, we can start on the solutions.
This album is powered by BubbleShare - Add to my blog
Now, moving forward we look into MAC’s approach; which has a similar reasoning but a different execution
This album is powered by BubbleShare - Add to my blog
Now moving on to Mr Alcantara’s approach, which uses a simpler reasoning on the problem
This album is powered by BubbleShare - Add to my blog
Now we are FINALLY done with the POD… it took me a while to do those ones.
I am now going to post about today’s ACTUAL lesson.
Sybil the rat : A mathematicla answer .
-Today we reviewed the problem of Sybil the rat : A mathematicla answer .
Sybil must 1st cross half of a room, the ¼ of it, then 1/8 and so on.
First thing we need to realize is, tn = fraction of room to cross. So when we are in doubt and don’t know what to do, we list the first terms of the sequence, which in this case are:
½ , ¼ , 1/8 , 1 /16 … as we could realize the explicit definition for this sequence is: n
(1/2)
>So we need to ask ourselves a question… what happens to tn as n approaches infinity?
First, we should give a graphical answer- REMEMBER that your grapgh shoud include
>A scale with labels
> tn values as it approaches its limit ( in this case zero)
> tn values as n approaches infinity
>must label both axes
>must show points not lines( remember it is a sequence)
>Interpret the graph
>Now what if we wanted to know the total amount of room crossed?
Sn= total amount of room crossed
If Sybil can cross the room then Sn> or equal to 1
List find first terms of the series.
Realize that it is a geometric series whichs r = 1/2
Well that was basically the most important part, Mr. Alcantara said something about using algebra on limits to cancel them out and get somenthing which I didn’t really understand. Mr. Alcantara said he did’t expect us to know it, so I’m not covering it now.
Homework was : Read 500-503
Pg. 502 1-11 odd
n
In the book they mention a new theorem, that if l r l < 1, then lim r = 0.
n->infinity
If you think about it it is totally logical because you are repeatedly multiplying by a number less than one, each time oits going to go closer to zero.
Well that’s it for today.
Monday, May 28, 2007
Frank and Nico's Scribe
Scribe.
1. P.O.D.
Larisa was being harrased by a mosquito. She picked up a fly swatter and wacked the moskito by snapping her wrist down but keeping the rest of her arm steady. Let us asume that the distance from Larisa's wrist to the head of the fly was 50cm. If the swatter swung through an anlge of 52º, how far did the head of the fly travel?
1. P.O.D.
Larisa was being harrased by a mosquito. She picked up a fly swatter and wacked the moskito by snapping her wrist down but keeping the rest of her arm steady. Let us asume that the distance from Larisa's wrist to the head of the fly was 50cm. If the swatter swung through an anlge of 52º, how far did the head of the fly travel?
- First, we need to realize that the segment traveled by the fly swatter is not a straight line, it's an arc.
- Then you need to remember past topics on our pre calculus class, for example: The Snowman units formula, and how to convert from degrees to radians and viceversa.
- Then you can tell that what we need to find out it the lenght of the arc.
In today's class, we recieved 3 answers from students in the class. First, Cristy Bustillo's answer. Unfortunately she didn't realize that the segment traveled by the fly swatter was an arc, not a line. So she drew a triangle and then used trig functions to find the lenght of the unknown side.
The second answer was given by MAC. She did realize that what we were looking for was an arc, so she used this formula: S = rƟ. S being the lenght of the arc, r being the radius, and Ɵ being the angle in radians. In order to use this formula, she had to convert the angle that she had in degrees to radians. She converted it and the plugged it in into the formula, to later get an answer for the lenght of the arc(S)
The third answer was given by Susy DC. She used proportions. She realized that 52 was a portion of the total angle in a circle, 360, and that S is a portion of the total circumference of the circle, 2πr. She also knew that both proportions were equal. Then she solved for x to the the same number that MAC got for her solution.
Here you can see the answers by our classmates and the procedure to get thereThis album is powered by BubbleShare - Add to my blog
After the P.O.D., that took pretty long, José Alcantara read us a story about a race between a tortoise and Achilles. which remembered us of the rat problem. Here you can find the story: www.mathacademy.com/pr/prime/articles/zeno_tort/ .
Think about it. Will Achilles get past the tortoise? in this case? in real life?
After reading the story, and discussing it, José Alcantara gave us some Limit practive problems so we get prepared for the next quiz!!
Limit Problems:
1. Solve without using your graphing calculator
2. Check your answers using the graphing calculator.
3. Check with José Alcantara's posted answersThis album is powered by BubbleShare - Add to my blogThe only homework was to correct the quizzes to get 5 point more.
And the next scribe has to be done by: Daniel Enrique Rubio Ferrer.
Thursday, May 24, 2007
Scribe Post
SCRIBE POST
In today’s Pre- Calculus lesson, we started off with our daily P.O.D. Throughout the class period we worked on a new subject; constants and on limits which we have been working on for a few days now.
P.O.D:
- Today’s P.O.D was divided into four parts:
a. List the first 10 terms of the Fibonacci sequence.
F1= 1 F2= 1 Fn= Fn-1 + Fn-2 - Recursive definition
Sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55…
b. Complete the table of values accurate to 4 decimals.
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c. Describe in words how the value ratio behaved as you completed the table. Be precise.
-As the nth term gets bigger the decimal place moves one place more.
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d. Approximate lim Fn , accurate to 3 decimals.
n-∞ Fn-1
- If we look at the table we did in part B for the later terms like:
tn 1.6190 1.6176 1.6181 1.6179
n 9 10 11 12
Once you round them to the thousands place, they all round up to 1.618 therefore
lim Fn = 1.618
nà∞ Fn-1
The ratio Fn oscillates above and below a constant and the constant approaches
Fn-1
as n approaches infinity.
-this constant is called Phi = Φ = 1.61803
-Φ is an irrational number just like π and е.
During class we tried to prove one of the constants that appeared in the movie the Da Vinci code. Allegedly the ratio of the length from your hip to the ground is the same to that of your total height. This is supposed to be 5.
-Pier volunteered for this activity
-Pier’s measurements: from hip to floor : 98 cm
Total height: 1”75 cm
1”75cm/ 98cm = 1.78 ratio
The next activity we did was a group exercise we had to:
On you index finger:
a. measure the length of the 3rd bone (knuckle to knuckle)
b. Measure the 2nd bone
c. Divide the longer by the shorter
d. Find the average on the board
e. Find Class Average.
My Group’s measurements:
Susy’s: 4cm,/2.5cm =1.6cm
Gaby’s: 4.8cm/2.8= 1.714 cm
Sofi’s: 4cm/2.5cm=1.8cm
group average: 1.703 cm
Class Average
-1.767
cm-1.6305
-1.6422
-1.6666
-1.6666
-1.703
Ans: 1.675 cm
Exercises:
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In today’s Pre- Calculus lesson, we started off with our daily P.O.D. Throughout the class period we worked on a new subject; constants and on limits which we have been working on for a few days now.
P.O.D:
- Today’s P.O.D was divided into four parts:
a. List the first 10 terms of the Fibonacci sequence.
F1= 1 F2= 1 Fn= Fn-1 + Fn-2 - Recursive definition
Sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55…
b. Complete the table of values accurate to 4 decimals.
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c. Describe in words how the value ratio behaved as you completed the table. Be precise.
-As the nth term gets bigger the decimal place moves one place more.
This album is powered by BubbleShare - Add to my blog
d. Approximate lim Fn , accurate to 3 decimals.
n-∞ Fn-1
- If we look at the table we did in part B for the later terms like:
tn 1.6190 1.6176 1.6181 1.6179
n 9 10 11 12
Once you round them to the thousands place, they all round up to 1.618 therefore
lim Fn = 1.618
nà∞ Fn-1
The ratio Fn oscillates above and below a constant and the constant approaches
Fn-1
as n approaches infinity.
-this constant is called Phi = Φ = 1.61803
-Φ is an irrational number just like π and е.
During class we tried to prove one of the constants that appeared in the movie the Da Vinci code. Allegedly the ratio of the length from your hip to the ground is the same to that of your total height. This is supposed to be 5.
-Pier volunteered for this activity
-Pier’s measurements: from hip to floor : 98 cm
Total height: 1”75 cm
1”75cm/ 98cm = 1.78 ratio
The next activity we did was a group exercise we had to:
On you index finger:
a. measure the length of the 3rd bone (knuckle to knuckle)
b. Measure the 2nd bone
c. Divide the longer by the shorter
d. Find the average on the board
e. Find Class Average.
My Group’s measurements:
Susy’s: 4cm,/2.5cm =1.6cm
Gaby’s: 4.8cm/2.8= 1.714 cm
Sofi’s: 4cm/2.5cm=1.8cm
group average: 1.703 cm
Class Average
-1.767
cm-1.6305
-1.6422
-1.6666
-1.6666
-1.703
Ans: 1.675 cm
Exercises:
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Tuesday, May 22, 2007
The correction of the correction of Mac and Nabil´s post
Today we continued to work on limits,by using graph and logic
-We began by solving the daily POD:
The east section of a stadium has 30 rows of seats the 10th row has 100 seats and every row has 3 more seats than the row above it. How many seats are in the east section of the stadium?
Analysis:
-First we identify what the problem is asking us to do: It is asking for the total amount of seats, so it is a series.
-It also says that each row has 3 more seats than the row above it,so the difference is 3 seats , so that makes it an arithmetic sequence.
-We are given T10that is 100and the r (ratio)is -3 because we are decreasing 3 seats from the of the term before.
- so first we find what T1 equals by using the explicit definition of an arithmetic sequence.
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- Now that we know what T1 is we can solve for T30:
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-Knowing what T1 and T30 are, we can solve for the sum of all rows using the formula for arithmetic series
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There are 2,205 seats in the east side of the stadium
-Continuing with the lesson on limits; We began studying how to find LIM without a calculator.
-There are two acceptable ways
A) list the first few terms
b)think
-the first problem we solved was:
Tn=(-n)^2 Lim Tn?
-Just by looking at the problem we can induce that with any N we plug in the value on Tn will be positive ( n is rise to an even power) then if we make a table of values for the first few terms, we realize that the outputs are increasing.
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Therefore the limit of Tn as n approaches infiniti, because Tn continues getting bigger.
The second problem was:
Tn=(-1)^n-1· n/10n Lim Tn?
-We made a table of values to find the tendency of Tn as n increases:
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-We can now state that the limit of Tn as n approaches to infiniti is 0, because the values of Tn keep getting closer to 0 as n increases.
The 3rd problem we solved was:
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-As we did in the last problem, we started with a tabla of values which is the following:
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- If we analize this way we can see that the value of Tn is not really approaching to one real number,it approaches both 1 and -1.This makes the limits DNE ,does not exist.
-The 4th problem we solevd was a little bit different because it asks for the limit of cosine.
Tn=cos(n) Lit tn=?
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And just like in problem #3, Tn is not approching 1 single real number, it approaches 1 and -1. the limit DNE .
-The last problem was:
Tn= Ln (n) Lim Tn= ?
-to solve this problem we have to know what the graph of LnX look like. We new how the graph of e^x was, and since LnX and e^x are apposites LnX is the reflection of e^x in the y=x axis
-Graph of e^x and y=x axis with it`s reflection
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- By seeing the graph, we can know that the limit of Tn as n approches infinity is infinity, because as n gets, Tn continues increasing as well
- We hope this was helpfull for you and that it cleared the doubts that you had. If you have any questions feel free to post them or add comments and we will be happy to answer them or explain.
!!!!!!!!!!!There is no¨NEW¨ homework.
- Next scribe Fiore and Caro I
-We began by solving the daily POD:
The east section of a stadium has 30 rows of seats the 10th row has 100 seats and every row has 3 more seats than the row above it. How many seats are in the east section of the stadium?
Analysis:
-First we identify what the problem is asking us to do: It is asking for the total amount of seats, so it is a series.
-It also says that each row has 3 more seats than the row above it,so the difference is 3 seats , so that makes it an arithmetic sequence.
-We are given T10that is 100and the r (ratio)is -3 because we are decreasing 3 seats from the of the term before.
- so first we find what T1 equals by using the explicit definition of an arithmetic sequence.
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- Now that we know what T1 is we can solve for T30:
This album is powered by BubbleShare - Add to my blog
-Knowing what T1 and T30 are, we can solve for the sum of all rows using the formula for arithmetic series
This album is powered by BubbleShare - Add to my blog
There are 2,205 seats in the east side of the stadium
-Continuing with the lesson on limits; We began studying how to find LIM without a calculator.
-There are two acceptable ways
A) list the first few terms
b)think
-the first problem we solved was:
Tn=(-n)^2 Lim Tn?
-Just by looking at the problem we can induce that with any N we plug in the value on Tn will be positive ( n is rise to an even power) then if we make a table of values for the first few terms, we realize that the outputs are increasing.
This album is powered by BubbleShare - Add to my blog
Therefore the limit of Tn as n approaches infiniti, because Tn continues getting bigger.
The second problem was:
Tn=(-1)^n-1· n/10n Lim Tn?
-We made a table of values to find the tendency of Tn as n increases:
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-We can now state that the limit of Tn as n approaches to infiniti is 0, because the values of Tn keep getting closer to 0 as n increases.
The 3rd problem we solved was:
This album is powered by BubbleShare - Add to my blog
-As we did in the last problem, we started with a tabla of values which is the following:
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- If we analize this way we can see that the value of Tn is not really approaching to one real number,it approaches both 1 and -1.This makes the limits DNE ,does not exist.
-The 4th problem we solevd was a little bit different because it asks for the limit of cosine.
Tn=cos(n) Lit tn=?
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And just like in problem #3, Tn is not approching 1 single real number, it approaches 1 and -1. the limit DNE .
-The last problem was:
Tn= Ln (n) Lim Tn= ?
-to solve this problem we have to know what the graph of LnX look like. We new how the graph of e^x was, and since LnX and e^x are apposites LnX is the reflection of e^x in the y=x axis
-Graph of e^x and y=x axis with it`s reflection
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- By seeing the graph, we can know that the limit of Tn as n approches infinity is infinity, because as n gets, Tn continues increasing as well
- We hope this was helpfull for you and that it cleared the doubts that you had. If you have any questions feel free to post them or add comments and we will be happy to answer them or explain.
!!!!!!!!!!!There is no¨NEW¨ homework.
- Next scribe Fiore and Caro I
Caro I and Fiore's Scribe (May 22)
Today in class we continued with the topic of Limits, and we learnes about one trick which makes it easier for us to solve the problems.
First we did our daily POD:
Tn = 1/n
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Some of these don’t really work for sequences. It’s better to change it as this kind of statement: F(X) = 1/x. This way you can think of it as a function.
Answers for the POD:
a. One way to know the answer to this question is by making a chart. Here you plug in any values, better if they are big numbers, for the variable n. Then in the equation Tn= 1/n replace n for each of the numbers to get Tn.
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In this case we notice that any huge number plugged in for n will make n and tn decrease in a way that it’s getting closer to 0.
So the limit of tn as n approaches to infinite is cero.
In a graph is would look like this:
- Here you can see how the dots plotted are tending to go near y= 0.
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The limit is 0, because is approaching to 0, every time a big number is plug in.
Both are negative, so in terms of a graph it would be on the 3rd quadrant.
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c.
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This numbers are not approaching infinity. They do not exist (DNE). This is because this problem could have two different limits and limits only exist when it’s approaching certain place by just one way.
(It is not important this year, in calculus we need to now a lot about one sided limits)
After the POD, we talked about a Limit Trick and did two problems.
(This is not on tomorrows quiz)
- the limit trick applioes only when X + or - infinite
- Multiply top and bottom by the reciprocal (divide) of the variable raised to the highest power in the denominator.
In the calculator you can prove this answer by plugging in 6x squared over (3x squared + 7x). The graph will be this one; which as you can notice, the only part that matters is as n goes to infinite. In this case n approaches to 2. y= 2.
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The second problem:
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Tomorrow Quiz will cover:
· Sequences
· Series
· Limits
In the scribes, many people were just posting information without any explanation. If you want to correct yours, look at Nabil and Mac’s scribe because they had a very nice explanation for people who really don’t understand the topic.
THEIR IS NO NEW HOMEWORK
- If you have any doubts or problem understanding our scribe post a comment or talk to us and we'll clear up everything. We hope you like it and that it is useful for your exams.
The next scribe must me done by SUSUAN KEQUERICA Y SOFIA NAVAS.
Monday, May 21, 2007
Limits-
Classmates I have good news for you .Look what I found=
http://en.wikipedia.org/wiki/Limit_(mathematics)
This page was really helpfull for me..hope it is for you.....
(It´s like a short lecture on limits.
http://en.wikipedia.org/wiki/Limit_(mathematics)
This page was really helpfull for me..hope it is for you.....
(It´s like a short lecture on limits.
Mac and Nabil´s sribe correction
As you realize..our graphs didnt came out .. so now Itried a different way.Ihope it comes out.(The graphs are from 1 to 9 in the last scribe the first one is nuber one in the correction the second is number two ,and so on.
1)
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2)
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3)
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4)
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5)
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6)
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7)
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8)
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9)
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1)
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2)
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3)
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4)
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5)
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6)
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7)
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8)
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9)
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Limits
Hey everyone:
I was searching on the internet for problems and more explanations of Limits so i can practice more about it. I found a page which helped me a lot to understand more about the topic. I wanted to share it with all of you.
i hope it can help you.
here is the page:
http://www.coolmath.com/limit1.htm
I was searching on the internet for problems and more explanations of Limits so i can practice more about it. I found a page which helped me a lot to understand more about the topic. I wanted to share it with all of you.
i hope it can help you.
here is the page:
http://www.coolmath.com/limit1.htm
Thursday, May 17, 2007
Mac and Nabil´s scribe
Today we continued to work on limits,by using graph and logic
-We began by solving the daily POD:
The east section of a stadium has 30 rows of seats the 10th row has 100 seats and every row has 3 more seats than the row above it. How many seats are in the east section of the stadium?
Analisis:
-First we identify what the problem is asking us to do: It is asking for the total amount of seats, so it is a series.
-It also sais that each row has 3 more seats than the row above it,so the difference is 3 seats , so that makes it an arithmetic sequence.
-We are given T10that is 100and the r (ratio)is -3 because we are decreasing 3 seats from the of the term before.
- so first we find what T1 equals by using the explicit definition of an arithmetic sequence.
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- Now that we know what T1 is we can solve for T30:
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-Knowing what T1 and T30 are, we can solve for the sum of all rows using the formula for arithmetic series
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-There are 2,205 seats in the east side of the stadium
-Continuing with the lesson on limits; We began studying how to find LIM without a calculator.
-There are two acceptable ways
A) list the first few terms
b)think
-the first problem we solved was:
Tn=(-n)^2 Lim Tn?
-Just by looking at the problem we can induce that with any N we plug in the value on Tn will be positive ( n is rise to an even power) then if we make a table of values for the first few terms, we realize that the outputs are increasing.
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Therefore the limit of Tn as n approaches infiniti, because Tn continues getting bigger.
The second problem was:
Tn=(-1)^n-1· n/10n Lim Tn?
-We made a table of values to find the tendency of Tn as n increases:
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-We can now state that the limit of Tn as n approaches to infiniti is 0, because the values of Tn keep getting closer to 0 as n increases.
The 3rd problem we solved was:
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-As we did in the last problem, we started with a tabla of values which is the following:
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- If we analizethis way we can see that the value of Tn is not really approaching to one real number,it approaches both 1 and -1.This makes the limits DNA ,does not exist.
-The 4th problem we solevd was a little bit different because it asks for the limit of cosine.
Tn=cos(n) Lit tn=?
- The graph of cos is :
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And just like in problem #3, Tn is not approching 1 single real number, it approaches 1 and -1. the limit DNE .
-The last problem was:
Tn= Ln (n) Lim Tn= ?
-to solve this problem we have to know what the graph of LnX look like. We new how the graph of e^x was, and since LnX and e^x are apposites LnX is the reflection of e^x in the y=x axis
-Graph of e^x and y=x axis with it`s reflection
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- By seeing the graph, we can know that the limit of Tn as n approches infinity is infinity, because as n gets, Tn continues increasing as well
- We hope this was helpfull for you and that it cleared the doubts that you had. If you have any questions feel free to post them or add comments and we will be happy to answer them or explain.
!!!!!!!!!!!There is no¨NEW¨ homework.
- Next scribe Fiore and Caro I
-We began by solving the daily POD:
The east section of a stadium has 30 rows of seats the 10th row has 100 seats and every row has 3 more seats than the row above it. How many seats are in the east section of the stadium?
Analisis:
-First we identify what the problem is asking us to do: It is asking for the total amount of seats, so it is a series.
-It also sais that each row has 3 more seats than the row above it,so the difference is 3 seats , so that makes it an arithmetic sequence.
-We are given T10that is 100and the r (ratio)is -3 because we are decreasing 3 seats from the of the term before.
- so first we find what T1 equals by using the explicit definition of an arithmetic sequence.
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- Now that we know what T1 is we can solve for T30:
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-Knowing what T1 and T30 are, we can solve for the sum of all rows using the formula for arithmetic series
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-There are 2,205 seats in the east side of the stadium
-Continuing with the lesson on limits; We began studying how to find LIM without a calculator.
-There are two acceptable ways
A) list the first few terms
b)think
-the first problem we solved was:
Tn=(-n)^2 Lim Tn?
-Just by looking at the problem we can induce that with any N we plug in the value on Tn will be positive ( n is rise to an even power) then if we make a table of values for the first few terms, we realize that the outputs are increasing.
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Therefore the limit of Tn as n approaches infiniti, because Tn continues getting bigger.
The second problem was:
Tn=(-1)^n-1· n/10n Lim Tn?
-We made a table of values to find the tendency of Tn as n increases:
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-We can now state that the limit of Tn as n approaches to infiniti is 0, because the values of Tn keep getting closer to 0 as n increases.
The 3rd problem we solved was:
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-As we did in the last problem, we started with a tabla of values which is the following:
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- If we analizethis way we can see that the value of Tn is not really approaching to one real number,it approaches both 1 and -1.This makes the limits DNA ,does not exist.
-The 4th problem we solevd was a little bit different because it asks for the limit of cosine.
Tn=cos(n) Lit tn=?
- The graph of cos is :
http://www.macromedia.com/go/getflashplayer" quality="high" src="http://www.bubbleshare.com/swfs/player.swf?3599" type="application/x-shockwave-flash" width="280">This album is powered by BubbleShare - Add to my blog
And just like in problem #3, Tn is not approching 1 single real number, it approaches 1 and -1. the limit DNE .
-The last problem was:
Tn= Ln (n) Lim Tn= ?
-to solve this problem we have to know what the graph of LnX look like. We new how the graph of e^x was, and since LnX and e^x are apposites LnX is the reflection of e^x in the y=x axis
-Graph of e^x and y=x axis with it`s reflection
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- By seeing the graph, we can know that the limit of Tn as n approches infinity is infinity, because as n gets, Tn continues increasing as well
- We hope this was helpfull for you and that it cleared the doubts that you had. If you have any questions feel free to post them or add comments and we will be happy to answer them or explain.
!!!!!!!!!!!There is no¨NEW¨ homework.
- Next scribe Fiore and Caro I
Wednesday, May 16, 2007
Jose and Chippy's Blog for Wednesday, May 16th, 07
Today's POD:
Mr. Alcantara was pushing his son Julian, who was in the swing. The arc he made was 3 meters long. When he stopped pushing him, each time the arc decreased 1/6. How man distance did Julian travel after 10 times of swinging?
We found out, that 1/6 of 3, is .5 Because of this, the next term is 2.5. Then, multiply that by 5/6 so we can get the third term, and so on. It is a geometric Sequence. It is good to always find the first few terms.
To find the distance Julian traveled, we plug in the numbers in the equation for the sum of the terms in a geometric sequence.
S10= 3(1-5/6^10) / 1- 5/6
The answer is: 15.09m
We were given a sequence:
1/2, 3/4, 7/8, 15/16
Find the explicit formula:
tn = 2^n-1/2^n
Then, we were asked to graph the sequence. Changes in the calculator were:
Under mode change:
-dot
-sequence graphing
y = nMin? y = nMin_1_
m(n) = 2^n-1/2^n
y = u(nMin)? y = u(nMin) = __1/2__...this gave us the graph of the sequence in the calculator.
This was to learn to use our calculators, but we didn't really need to know this. And, to understand better the term "limit", seeing it in the calculator.
Next scribe: Mac and Nabil
Mr. Alcantara was pushing his son Julian, who was in the swing. The arc he made was 3 meters long. When he stopped pushing him, each time the arc decreased 1/6. How man distance did Julian travel after 10 times of swinging?
We found out, that 1/6 of 3, is .5 Because of this, the next term is 2.5. Then, multiply that by 5/6 so we can get the third term, and so on. It is a geometric Sequence. It is good to always find the first few terms.
To find the distance Julian traveled, we plug in the numbers in the equation for the sum of the terms in a geometric sequence.
S10= 3(1-5/6^10) / 1- 5/6
The answer is: 15.09m
We were given a sequence:
1/2, 3/4, 7/8, 15/16
Find the explicit formula:
tn = 2^n-1/2^n
Then, we were asked to graph the sequence. Changes in the calculator were:
Under mode change:
-dot
-sequence graphing
y = nMin? y = nMin_1_
m(n) = 2^n-1/2^n
y = u(nMin)? y = u(nMin) = __1/2__...this gave us the graph of the sequence in the calculator.
This was to learn to use our calculators, but we didn't really need to know this. And, to understand better the term "limit", seeing it in the calculator.
Next scribe: Mac and Nabil
Jose and Chippy's Scribe for Tuesday, May 15th, 07
The POD for today was:
S is a geometric sequence. T3= -7 T6= -2401
Find t10
What we have to do, in order to solve the problem is to find the rate. To do so, we plug in the equation t3 as t1.
-2401= -7*r^3
We use three as the exponent because there are three jumps from t3 to t6.
Solving for r, me stay with this:
√(r^3)= √(343) - Then, we cancel the exponent on r, cubing the root of 343. The cube root of 343 is 7.
After finding the rate, we can find t10. Knowing that we used t3 as t1, we can use t8 as t10 to find the answer to the problem.
t8= -7 * (7)^(8-1)
t8= -7 * (7)^7
t8= -5, 764,801, which is t10
After finishing the POD, we started with infinite sequences.
Infinite Sequence: A sequence that goes on forever. They may or may not have finite limits.
Another way to think of a limit, is to think of it as a boundary. Some can be reached, and some can't be reached, but you can't exceed it.
The bouncing ball example
The bouncing ball example had a limit. 100,50,25,...
As n gets lower, what is the value that tn is approaching?
" The limit of tn as n approaches infinity is 0"
Limtn = 0
n-->OO
Limtn= tells you to find the limit of a sequence.
Example 1: The sequence tn=n
1,2,3,4,5,... the limit is Limtn= 0
n-->OO
If a limit exists, it has to be a real number.
The Limit Does Not Exists. (DNE)
Limtn= 0
n-->OO
Example 2: -1, -4. -9, -16
What is the limit? Answer: DNE..it will keep going on to negative infinity. It also can be:
Limtn= - OO
n-->OO
Example 3: There are grains of salt in an infitnite chess board.
2,4,8...
What is the Limit?
Limtn= -OO
n--> OO
Example 4:
1/2, 3/4, 7/8, 15/16,...
What is the limit?
As you can see, the number is approaching to 1, but it never gets to 1. So the answer is:
Limtn: 1
n-->OO
Example 5: Find Limit, if tn =(.99)^n
The limit is 0. Each time you get .99 of the number so it decreases, but never reaches 0.
The safest way to do it:
-Plug large values of n
-List terms
-Repeated multiplication of a number less than 1.
-99%
S is a geometric sequence. T3= -7 T6= -2401
Find t10
What we have to do, in order to solve the problem is to find the rate. To do so, we plug in the equation t3 as t1.
-2401= -7*r^3
We use three as the exponent because there are three jumps from t3 to t6.
Solving for r, me stay with this:
√(r^3)= √(343) - Then, we cancel the exponent on r, cubing the root of 343. The cube root of 343 is 7.
After finding the rate, we can find t10. Knowing that we used t3 as t1, we can use t8 as t10 to find the answer to the problem.
t8= -7 * (7)^(8-1)
t8= -7 * (7)^7
t8= -5, 764,801, which is t10
After finishing the POD, we started with infinite sequences.
Infinite Sequence: A sequence that goes on forever. They may or may not have finite limits.
Another way to think of a limit, is to think of it as a boundary. Some can be reached, and some can't be reached, but you can't exceed it.
The bouncing ball example
The bouncing ball example had a limit. 100,50,25,...
As n gets lower, what is the value that tn is approaching?
" The limit of tn as n approaches infinity is 0"
Limtn = 0
n-->OO
Limtn= tells you to find the limit of a sequence.
Example 1: The sequence tn=n
1,2,3,4,5,... the limit is Limtn= 0
n-->OO
If a limit exists, it has to be a real number.
The Limit Does Not Exists. (DNE)
Limtn= 0
n-->OO
Example 2: -1, -4. -9, -16
What is the limit? Answer: DNE..it will keep going on to negative infinity. It also can be:
Limtn= - OO
n-->OO
Example 3: There are grains of salt in an infitnite chess board.
2,4,8...
What is the Limit?
Limtn= -OO
n--> OO
Example 4:
1/2, 3/4, 7/8, 15/16,...
What is the limit?
As you can see, the number is approaching to 1, but it never gets to 1. So the answer is:
Limtn: 1
n-->OO
Example 5: Find Limit, if tn =(.99)^n
The limit is 0. Each time you get .99 of the number so it decreases, but never reaches 0.
The safest way to do it:
-Plug large values of n
-List terms
-Repeated multiplication of a number less than 1.
-99%
Tuesday, May 15, 2007
Sybil the Rat
The rat, Sibyl, who lives beneath the trailer, wants to cross the room from one end to the other. Before she can cross the full distance, she must cross half the distance. Before she can cross the rest (half of the full distance) she must cross the next ¼. Before she can cover the last ¼, she must cover the next 1/8. And so on. Since covering any distance requires some time, it would take Sybil an infinite amount of time to cross all of the infinite distances. Therefore, Sibyl can never cross the room.
By the same argument, you could never cross the room either.
How can this be?
By the same argument, you could never cross the room either.
How can this be?
Monday, May 14, 2007
CHEWY'S SCRIBE POST
Hi everyone! It’s me Chewy, today is my turn to scribe my post. We started class with a usual POD, today’s problem was:
POD 5/14:
In 1990, 2,509 surfboards were made in Brazil. Statistics show that each year the production increases by 659 surfboards. According to statistics how many surfboards have made from 1990 to the end of last year?
As soon as we read the problem we deduce that is an arithmetic series problem.
The question is how can we solve it? First we have to subtract the last year to the first, after doing the math we noticed that 2006-1990= 16, but we have to add 1 because there were surfboards also made in year 1990, so it equals to 17. Then to find the amount of surfboards that were made from 1990 to the end of last year, we have to use a formula, which is: Tn= T1+(n-1)d. Then we plug in T1 which is 2,509 and n which is 17, and d is 659. Then plug in all does numbers in the formula.
Procedure:
Tn= T1+(n-1)d
T17=2,509+(17-1)(659)
T17=2,509+(16)(659)
T17=2,509+10544
T16= 13,053
S17= 17/2(2,509+13,053)
S17=132,277
Then the class continued and we start to practice some arithmetic and geometric series problems with the word problems that 11B did for homework.
Some examples are:
a) A special machine gun shoots 10 bullets per second, but each second its rate doubles. So if in the 1st second it shoots 10 bullets then in the 2nd one it’s going to shoot 20, n the 3rd 40 and so on. Find how many bullets can be shot in 60 seconds.
Immediately we noticed that this is an geometric series problem.
1second= 10 bullets
2second=20 bullets
3seconds=40bullets
We noticed that each second the bullet doubles. So we have to use this formula in order to get the answer. Sn=T1(1-r^n)/(1-r). We know that T1= 10, n=60 and r=2 because each time it doubles.
Procedure:
S60= 10(1-2^60)/9(1-2)
S60= 1.1529215*10^19 bullets.
b) One student fell on the soccer field 2 friends came to see what happened, every minute 2 more students came, than the one’s that came before. In 10 minutes how many students will there be on the soccer field?
This is an Arithmetic Series problem.
In order to solve this problem we have to use the next equation: Tn=T1+(n-1)d.
So T1 is 3 because in minute 1 there are 2 friends that arrive plus the one that is injured, n is 10 because we are looking how many students will there be on the soccer field in 10 minutes and d is 2 because the sequence adds 2 each time.
Procedure:
T10= 3+(10-1)(2)
T10=3+(9)(2)
T10=3+18
T10=21 students.
c) A skater started with a velocity of 4 km/hr, if he doubles his velocity each kilometer what will be his velocity on the 30th kilometer?
This is a Geometric Series problem.
We noticed that in 1km=4km/hr, 2km=8km/hr and 3km=16km/hr. So in order to get the correct answer we should use the next formula: Tn=T1*r^n-1.
Tn= 30 because we are looking for the velocity in 30km.
T1= 4
r=2
Procedure:
T30=4*2^30-1
T30=4*2^29
T30=4*536,870,912
T30=2,147,483,648
POD 5/14:
In 1990, 2,509 surfboards were made in Brazil. Statistics show that each year the production increases by 659 surfboards. According to statistics how many surfboards have made from 1990 to the end of last year?
As soon as we read the problem we deduce that is an arithmetic series problem.
The question is how can we solve it? First we have to subtract the last year to the first, after doing the math we noticed that 2006-1990= 16, but we have to add 1 because there were surfboards also made in year 1990, so it equals to 17. Then to find the amount of surfboards that were made from 1990 to the end of last year, we have to use a formula, which is: Tn= T1+(n-1)d. Then we plug in T1 which is 2,509 and n which is 17, and d is 659. Then plug in all does numbers in the formula.
Procedure:
Tn= T1+(n-1)d
T17=2,509+(17-1)(659)
T17=2,509+(16)(659)
T17=2,509+10544
T16= 13,053
S17= 17/2(2,509+13,053)
S17=132,277
Then the class continued and we start to practice some arithmetic and geometric series problems with the word problems that 11B did for homework.
Some examples are:
a) A special machine gun shoots 10 bullets per second, but each second its rate doubles. So if in the 1st second it shoots 10 bullets then in the 2nd one it’s going to shoot 20, n the 3rd 40 and so on. Find how many bullets can be shot in 60 seconds.
Immediately we noticed that this is an geometric series problem.
1second= 10 bullets
2second=20 bullets
3seconds=40bullets
We noticed that each second the bullet doubles. So we have to use this formula in order to get the answer. Sn=T1(1-r^n)/(1-r). We know that T1= 10, n=60 and r=2 because each time it doubles.
Procedure:
S60= 10(1-2^60)/9(1-2)
S60= 1.1529215*10^19 bullets.
b) One student fell on the soccer field 2 friends came to see what happened, every minute 2 more students came, than the one’s that came before. In 10 minutes how many students will there be on the soccer field?
This is an Arithmetic Series problem.
In order to solve this problem we have to use the next equation: Tn=T1+(n-1)d.
So T1 is 3 because in minute 1 there are 2 friends that arrive plus the one that is injured, n is 10 because we are looking how many students will there be on the soccer field in 10 minutes and d is 2 because the sequence adds 2 each time.
Procedure:
T10= 3+(10-1)(2)
T10=3+(9)(2)
T10=3+18
T10=21 students.
c) A skater started with a velocity of 4 km/hr, if he doubles his velocity each kilometer what will be his velocity on the 30th kilometer?
This is a Geometric Series problem.
We noticed that in 1km=4km/hr, 2km=8km/hr and 3km=16km/hr. So in order to get the correct answer we should use the next formula: Tn=T1*r^n-1.
Tn= 30 because we are looking for the velocity in 30km.
T1= 4
r=2
Procedure:
T30=4*2^30-1
T30=4*2^29
T30=4*536,870,912
T30=2,147,483,648
THIS IS MY SCRIBE POST, I HOCHPE IT HELP YOU GUYS TO ANSWER ALL YOUR DOUBTS AND QUESTIONS.
cheeewwwyyy...
***I choose chippy and jose to do the next scribe.***
Scribe Post grade question
Mr. Alcantara,
I have a question regarding the scribe post.
The grade we receive when we do it, is it a part of our blog grade, or is it a completely different grade?
I was talking to some people in class, and some of 11B, and they also had the same question. Thanks!
I have a question regarding the scribe post.
The grade we receive when we do it, is it a part of our blog grade, or is it a completely different grade?
I was talking to some people in class, and some of 11B, and they also had the same question. Thanks!
Friday, May 11, 2007
SUSY LEC, SOFY, AND GABYS POST.
1. You visit the Great Pyramid and drop a penny off the tip of it. The distance the penny will fall is 16 feet the first second, 48 feet the next second, 80 feet the third second. What is the total distance the object will fall in 6 seconds?
2.
If a ball is initially dropped from a height
of 6 meters and bounces 57% of the height of its previous
fall, find the total vertical distance traveled by
the ball.
2.
If a ball is initially dropped from a height
of 6 meters and bounces 57% of the height of its previous
fall, find the total vertical distance traveled by
the ball.
Word Problems! By: MAC, Caro Ibanez and Nabily
1) A snake produces 2 snakes in 1 month. Each month the 2 snakes produce 2 snakes more each. In the 2nd month there are 6 snakes. So how many snakes there will be in 1 year and 2 months?(pretend all the snakes reproduce snakes, they are all girls)
2) Colegio Jorge Washington accepts 8 new students for each class every year, if a class begins with 15 students in pre-kinder,
A) how many students will there be when that class is in 7th grade
B) By the time they graduate (Assuming that no one failed a year or dropped out of school)
2) Colegio Jorge Washington accepts 8 new students for each class every year, if a class begins with 15 students in pre-kinder,
A) how many students will there be when that class is in 7th grade
B) By the time they graduate (Assuming that no one failed a year or dropped out of school)
Word Problems... Pier, Cristy B, Chewy
1. There is a rumor going around. On the first day 10 people find out, five days later 160 people find out about the rumor. How many people will know the rumor by day 10?
2. Juliana loves books. Every Friday she buys books. The first Friday she bought 3 books, and from that Friday on, every time she bought books she bought three more than the ones she had bought the week before. How many books would she have bought in total after a year of buying books?
2. Juliana loves books. Every Friday she buys books. The first Friday she bought 3 books, and from that Friday on, every time she bought books she bought three more than the ones she had bought the week before. How many books would she have bought in total after a year of buying books?
Scribe May 11/2007
Today, we were greeted, like everyday, with a POD! Today's problem was: a) Trace your direct ancestors back ten generations. How many people does that include? Ignore any inbreeding. b) How many generations must you go back to find 1 million ancestors.
Answers:
a) 2 ( 1 - 2^10/1-2) = 2046 We get this answer by using the geometric series formula, which in this class applies, because each generation we are multiplying the number of ancestors we have by two. When we divide the number of grandparents we have (4) by the number of parents we have (2), we get 2, so R = 2. N in this formula is equal to 10 because we are looking for how many ancestors we have ten generations back.
b) 1,000,000 = 2 [ 1-(2^n)/1-2]
500,000 = [1-(2^n)/1-2]
-500,000 = 1- (2^n)
-500,001 = (-2)^n
500,001 = 2^n Log2
500,001 = Log2 2^n
Log2 500,001 = n
Log 500,001/ Log 2 = 18.93 generations, or 19 generations.
In this part of the problem we are using a geometric series formula. Because we know Sn = 1,000,000, we can plug it in right away. Then, we figure out that T1 is equal to 2, because our first two direct ancestors are our parents. We know have two of the three variables that we can plug into this equation, so we are ready to solve for the third one, N. The algebra is pretty straight-forward, but, at the end of the problem we face a challenge, that most people in class were not able to figure out.Log2 500,001 = n... what do we do now?? Mr. Alcantara refreshed our memories, and showed us that:
if Log2 500,001 = y
then 2^y = 500,001
So, we solve now... Log 2^y = Log 500,001
y Log 2 = Log 500,001
y Log 2/Log 2 = Log 500,001/Log 2
y= Log 500,001/Log 2 y= 18.93
After explaining this to us, we agreed that there is no possible way to have 18.93 generations, you either have 18 or 19 generations, so, we decided, that to have one million direct ancestors, you must at least have 19 generations.
We then moved on into practice problems We had started to do these in class on Thursday, but we didn’t get to finish them, so we checked our answers today. The problem was: If T1 is = 1 and if T2 is = -3 Find S20
A) if the series is geometric b) if the series is arithmetic
a) The first thing you have to do is find R
r= T2/T1
r= -3/1
r= -3
Now, we can find S20.
S20 = 1 [1 – (-3)^20 / 1 – (-3)]
S20 = -871, 696, 100 .5
b) In b, the first thing you have to do is find d.
d = T2 –T1d= -3 – 1
d= -4
Now, we need to find T20, so that we can plug it into the arithmetic sequences equation.
T20 = 1 + (20-1) x -4
T20 = 1+ 19(-4)
T20 = 1 + -76
T20 = -75
Now, we have everything we need to plug into the arithmetic sequences equation.
S20 = (20/2) (1-75)
S20 = 10 x -74
S20 = -740
The second practice problem was: A grain of salt is placed on the corner square of a large chess board. Two grains are placed on the next square, four grains on the next, then eight, and so on. a) How many grains will the board have when half the squares have been covered? b) B) When all the squares have been covered? · A chessboard has 64 squares!
Answers:
a) S32 = 1 [1- (2^32) / 1-2]
S32 = 4,294,967,295 grains
We get his answer because r = 2 and we are looking for half the chess board, if it has 64 squares, then half the squares are 32 squares. We plug this information into a geometric sequences equation and we get the answer.
b) S64 = 1 [1- (2 ^64) / 1-2]
S64 = 1.8446744 X 10^19 grains.
We get this answer using the same information as in part a, only that we change 32 for 64 because we are looking for the total of the chessboard. After that, we were asked a new question using the same word problem…
c) How many grains were placed on the 16th square. We must realize that this is a geometric series question, it is not a geometric sequence, like the questions above.
T16 = 1 x 2^(16-1)
T16 = 1 (2^15) T16 = 2^15
T16 = 32, 768
We use the arithmetic series formula, because we only want the number of grains in square number 16, not the sum of the grains up to and including square 16.
The homework on the board is Pg 489 # 9-14, 17-20, 27, 29, 35 and, we must also post on the blog two word problems, one has to be an arithmetic series, and the other one a geometric series. Please post these problems without the solutions.
The next scribe will be… Chewy… who is alone… Have a nice weekend!! :D
Posted by: Cristina B and Pier
Answers:
a) 2 ( 1 - 2^10/1-2) = 2046 We get this answer by using the geometric series formula, which in this class applies, because each generation we are multiplying the number of ancestors we have by two. When we divide the number of grandparents we have (4) by the number of parents we have (2), we get 2, so R = 2. N in this formula is equal to 10 because we are looking for how many ancestors we have ten generations back.
b) 1,000,000 = 2 [ 1-(2^n)/1-2]
500,000 = [1-(2^n)/1-2]
-500,000 = 1- (2^n)
-500,001 = (-2)^n
500,001 = 2^n Log2
500,001 = Log2 2^n
Log2 500,001 = n
Log 500,001/ Log 2 = 18.93 generations, or 19 generations.
In this part of the problem we are using a geometric series formula. Because we know Sn = 1,000,000, we can plug it in right away. Then, we figure out that T1 is equal to 2, because our first two direct ancestors are our parents. We know have two of the three variables that we can plug into this equation, so we are ready to solve for the third one, N. The algebra is pretty straight-forward, but, at the end of the problem we face a challenge, that most people in class were not able to figure out.Log2 500,001 = n... what do we do now?? Mr. Alcantara refreshed our memories, and showed us that:
if Log2 500,001 = y
then 2^y = 500,001
So, we solve now... Log 2^y = Log 500,001
y Log 2 = Log 500,001
y Log 2/Log 2 = Log 500,001/Log 2
y= Log 500,001/Log 2 y= 18.93
After explaining this to us, we agreed that there is no possible way to have 18.93 generations, you either have 18 or 19 generations, so, we decided, that to have one million direct ancestors, you must at least have 19 generations.
We then moved on into practice problems We had started to do these in class on Thursday, but we didn’t get to finish them, so we checked our answers today. The problem was: If T1 is = 1 and if T2 is = -3 Find S20
A) if the series is geometric b) if the series is arithmetic
a) The first thing you have to do is find R
r= T2/T1
r= -3/1
r= -3
Now, we can find S20.
S20 = 1 [1 – (-3)^20 / 1 – (-3)]
S20 = -871, 696, 100 .5
b) In b, the first thing you have to do is find d.
d = T2 –T1d= -3 – 1
d= -4
Now, we need to find T20, so that we can plug it into the arithmetic sequences equation.
T20 = 1 + (20-1) x -4
T20 = 1+ 19(-4)
T20 = 1 + -76
T20 = -75
Now, we have everything we need to plug into the arithmetic sequences equation.
S20 = (20/2) (1-75)
S20 = 10 x -74
S20 = -740
The second practice problem was: A grain of salt is placed on the corner square of a large chess board. Two grains are placed on the next square, four grains on the next, then eight, and so on. a) How many grains will the board have when half the squares have been covered? b) B) When all the squares have been covered? · A chessboard has 64 squares!
Answers:
a) S32 = 1 [1- (2^32) / 1-2]
S32 = 4,294,967,295 grains
We get his answer because r = 2 and we are looking for half the chess board, if it has 64 squares, then half the squares are 32 squares. We plug this information into a geometric sequences equation and we get the answer.
b) S64 = 1 [1- (2 ^64) / 1-2]
S64 = 1.8446744 X 10^19 grains.
We get this answer using the same information as in part a, only that we change 32 for 64 because we are looking for the total of the chessboard. After that, we were asked a new question using the same word problem…
c) How many grains were placed on the 16th square. We must realize that this is a geometric series question, it is not a geometric sequence, like the questions above.
T16 = 1 x 2^(16-1)
T16 = 1 (2^15) T16 = 2^15
T16 = 32, 768
We use the arithmetic series formula, because we only want the number of grains in square number 16, not the sum of the grains up to and including square 16.
The homework on the board is Pg 489 # 9-14, 17-20, 27, 29, 35 and, we must also post on the blog two word problems, one has to be an arithmetic series, and the other one a geometric series. Please post these problems without the solutions.
The next scribe will be… Chewy… who is alone… Have a nice weekend!! :D
Posted by: Cristina B and Pier
Geometric & Arithmetic series: Word Problems By:Daniel Rubio, Diego Canelos and Abraham Farah.
1. Your best friend told you a secret today and you want to tell 2 more people. On day 2 you tell those 2 people. By the end of the second day you and your 2 friends know the secret.If each person tells 2 friends the secret, how many people would know the secret( excluding your best friend) by the end of day 9?
2.Your house has been invaded by cockraches and you need to know how many cockroaches there are. There are 3 cockroaches on day 1 , and every day 3 more cockroaches than the day before come to your house, find how many cockraches would there be by the end of day 9.
2.Your house has been invaded by cockraches and you need to know how many cockroaches there are. There are 3 cockroaches on day 1 , and every day 3 more cockroaches than the day before come to your house, find how many cockraches would there be by the end of day 9.
Chippy and Jose's word problems
1) Arithmetic Series
Antonio went to the doctor. For his growth, the doctor recommended that he should drink three glasses of milk each day, and each day drink half more glass than the last day. Find the total amount of glasses he drank by 2 months. B) Find how many glasses he drank at day 42.
2) Sandra bought three dogs in the store. She was a dog addict, and she had to have more and more dogs each week. The next week she had nine dogs in her house. After two weeks, her neighbor found 27 dogs. Find out how many dogs she had after 27 weeks. B) If she had started out with one dog, how many dogs would she have in the 27th week?
Antonio went to the doctor. For his growth, the doctor recommended that he should drink three glasses of milk each day, and each day drink half more glass than the last day. Find the total amount of glasses he drank by 2 months. B) Find how many glasses he drank at day 42.
2) Sandra bought three dogs in the store. She was a dog addict, and she had to have more and more dogs each week. The next week she had nine dogs in her house. After two weeks, her neighbor found 27 dogs. Find out how many dogs she had after 27 weeks. B) If she had started out with one dog, how many dogs would she have in the 27th week?
Word Problems by Mau, Lala and Bee
- There is a basket filled with frogs. A student gets the basket and tries to play a prank on the rest of the students. He goes to a classroom and places 3 frogs in it. Then, he goes to the second classroom and places 3 more frogs than he did in the previous classroom. In the third classroom he places 3 more than in the second, and so on until he puts in the 15th and last classroom 45 frogs. What is the total amount of frogs in all the classes?
- Susana goes to the store and buys one chocolate bar. She ate it and love it so much that she went back to the store and bought 4 chocolate bars. Then, she went back to the store and bought 16 chocolate bars. If she went 16 times to the store, how many chocolate bars did she buy?
Word Problems by Nico, Frank and DC
1. Every day, Frank eats 1 burger more than the day before. Each burger weighs 500g. At the end of a leap year, how many kilograms would frank have gained if on the first day of the year he ate 1 burger?
Assume he does not have a digestive system.
2. A young athlete has decided to run everyday as a rutine workout. He has a tournament 7 weeks from now. He decides to start easy and get harder as each week passes. He starts running 200 ft the first day and does that for a week. The next week he starts running 350ft and does that for a week. How many feet would he have run in total by the start of the tournament?
Assume he does not have a digestive system.
2. A young athlete has decided to run everyday as a rutine workout. He has a tournament 7 weeks from now. He decides to start easy and get harder as each week passes. He starts running 200 ft the first day and does that for a week. The next week he starts running 350ft and does that for a week. How many feet would he have run in total by the start of the tournament?
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