It seems like yesterday we have started our journey.
It was my pleasure having you as my students.
I wish you all the best during final exam (February 1, 2012 9:00 AM-11:00 AM in the gym)
and in your future endeavors.
Take care.
Mr.P
Friday, January 27, 2012
Thursday, January 19, 2012
Hello My name is Ryley and i well be your scribe for today.
before we start my computer doesn't like using the greater than (>),less than(<) signs so when you see < or > they mean < = less than
> = greater than
Infinite Geometric Series
Terms to know
Geometric sequence
-a geometric sequence is a sequence where each term aster the first is found by multiplying the previous term by a common ratio.
An example could be 2, 4, 8, 16, 32 ...., 131072 where the common ratio is 2.
Infinite Sequence
-contains an endless amount of numbers in the domain.
An example could be 3, 9, 27, 81,........ with no final term, it is infinite.
Infinite Geometric Series
-is the sum of all terms in an Infinite Geometric Sequence
-if |r| < 1 , then we have a convergent series, this means that the terms are getting smaller which also means that you can calculate a sum.
-if |r| > 1 , then we have a divergent series, this means that the terms in the sequence are becoming larger and you cannot calculate the sum of the terms.
The Infinite Geometric Series Formula is.
to be able to use this formula the sequence must be convergent (|r|< 1).
Example
36 + 6 + 1 + ......
to find "r" take the second term and divide it by the first term
t2
t1
because |r| is less than 1 you can find the sum.
using this formula find the sum.
Subtract the denominators by making them the same
next bring the denominators on top and cross multiply. (they have already been flipped)
which gives you...
So that's how to find the sum in a infinite Geometric series.
before we start my computer doesn't like using the greater than (>),less than(<) signs so when you see < or > they mean < = less than
> = greater than
Infinite Geometric Series
Terms to know
Geometric sequence
-a geometric sequence is a sequence where each term aster the first is found by multiplying the previous term by a common ratio.
An example could be 2, 4, 8, 16, 32 ...., 131072 where the common ratio is 2.
Infinite Sequence
-contains an endless amount of numbers in the domain.
An example could be 3, 9, 27, 81,........ with no final term, it is infinite.
Infinite Geometric Series
-is the sum of all terms in an Infinite Geometric Sequence
-if |r| < 1 , then we have a convergent series, this means that the terms are getting smaller which also means that you can calculate a sum.
-if |r| > 1 , then we have a divergent series, this means that the terms in the sequence are becoming larger and you cannot calculate the sum of the terms.
The Infinite Geometric Series Formula is.
to be able to use this formula the sequence must be convergent (|r|< 1).Example
36 + 6 + 1 + ......
to find "r" take the second term and divide it by the first term
t2
t1because |r| is less than 1 you can find the sum.
using this formula find the sum.
Subtract the denominators by making them the same
next bring the denominators on top and cross multiply. (they have already been flipped)
which gives you...
So that's how to find the sum in a infinite Geometric series.
Wednesday, January 18, 2012
Geometric Series
Hi Everyone, this is Vivian blogging again!! I am going to be explaining Geometric Series.
Series:
1.) The series of a geometric sequence is the sum of certain number of terms in the sequence or the sum of all of the terms in the sequence.
2.) To find the sum of n terms in a geometric sequence use the formula:
Sn = t1 (1-r^n) / 1-r
Sigma Notation:
1.) Is an alternate notation for finding geometric series (the sum)
2.) Sigma means the sum of all terms in a sequence.
Formulas for Geometric Sequences:
1.) Common Ration r = t2/t1 Note: r = t2/t1 = t3/t2 = t4/t3 = t5/t4, etc.
3.) To find one geometric mean use: +- √ (t1)(t3)
4.) When finding more than one geometric mean use tn = t1 x r^n-1 to find the common ratio and then multiply to find the missing terms.
Example 1:
Given the geometric sequence 2,6,18........find the sum of the first ten terms.
1.) First thing is you list out what you know.
t1 = 2
r = 6/2 = 3
2.) Then you solve for S10, so you use the formula Sn = t1 (1-r^n) / 1-r.
3.) You then plug in the numbers into your formula so:
S10 = 2 (1-3^10)/1-3
4.) After calculating this out your answer for S10 = 59048.
Example 2:
Find the sum of the first fifteen terms for a geometric sequence whose first term is 4920 and whose common ratio is 1/2.
1.) List out what you know:
t1 = 4820, r = 1/2
2.) You're solving for the sum of the first fifteen terms so the formula is Sn = t1 (1-r^n) / 1-r.
3.) Plug your numbers into the formula:
S15 = 4820 (1 - 1/2^15) / 1-1/2
4.) S15 = 9638.705811.
Hope this helped at least one person! I'm not good at blogging so I don't have pictures. Bye!! Have a nice day :)
Series:
1.) The series of a geometric sequence is the sum of certain number of terms in the sequence or the sum of all of the terms in the sequence.
2.) To find the sum of n terms in a geometric sequence use the formula:
Sn = t1 (1-r^n) / 1-r
Sigma Notation:
1.) Is an alternate notation for finding geometric series (the sum)
2.) Sigma means the sum of all terms in a sequence.
Formulas for Geometric Sequences:
1.) Common Ration r = t2/t1 Note: r = t2/t1 = t3/t2 = t4/t3 = t5/t4, etc.
2.) To find a specific term use the formula: tn = t1 x r^n-1
3.) To find one geometric mean use: +- √ (t1)(t3)
4.) When finding more than one geometric mean use tn = t1 x r^n-1 to find the common ratio and then multiply to find the missing terms.
Example 1:
Given the geometric sequence 2,6,18........find the sum of the first ten terms.
1.) First thing is you list out what you know.
t1 = 2
r = 6/2 = 3
2.) Then you solve for S10, so you use the formula Sn = t1 (1-r^n) / 1-r.
3.) You then plug in the numbers into your formula so:
S10 = 2 (1-3^10)/1-3
4.) After calculating this out your answer for S10 = 59048.
Example 2:
Find the sum of the first fifteen terms for a geometric sequence whose first term is 4920 and whose common ratio is 1/2.
1.) List out what you know:
t1 = 4820, r = 1/2
2.) You're solving for the sum of the first fifteen terms so the formula is Sn = t1 (1-r^n) / 1-r.
3.) Plug your numbers into the formula:
S15 = 4820 (1 - 1/2^15) / 1-1/2
4.) S15 = 9638.705811.
Hope this helped at least one person! I'm not good at blogging so I don't have pictures. Bye!! Have a nice day :)
Tuesday, January 17, 2012
Geometric Sequence
Hi! This is Phoebe and today we we learned about Geometric Sequence
Geometric Sequence.
Sequence
- is a function whose domain is set of consecutive positive intergers adn whose raance is an element of real numbers.
- It is also formed by a following rule, formula, pattern or equation
- It can be either finite or infinite
- Contains a limited number of elements in the domain. (This means it has an ending and a beginning)
- Contains and unlimited number of elements in the domain. (As a beginning but is never ending)
- A sequence in which each term, after the first, is found by multiplying previous term by a common ratio r.
- For me the main difference between Arithmetic Sequences and Geometric Sequences is that Arithmetic is more on Addition while Geometric is more on multiplication
- Common ration: r = t2/t1
ex. 6, 36, 216, 1296
36/6 = 6
- To find specific term use: tn = t1 x r^n-1
find the 12th term
t^12 = 6 x 6^12-1
t^12 = 6 x 6^11
t^12 = 2,176, 782, 336
- To find one geometric mean use: ± √t1 x t3
find the Geometric mean
± √6 x 216
± 36
- To find more than one geometric mean use tn = t1 x r^n-1 to find the common ratio and then multiply to find the missing term
Monday, January 16, 2012
Geometric Sequences & Series
Sorry it's taken me so long to post this blog entry!! Lets recap what has happened over the past couple of days..
Hopefully everyone was happy with their test results, had a great weekend, and finished the assignment that was due. Also, Happy belated "name day" to Mr. P's mom!
Oh yeah! We also stared learning a new unit.
Geometric Sequences & Series:
Lets start with the basics;
Example: Find the next term in each sequence. Describe how you got them.
1, 4, 9, 16, .... The answer is 25. We get this by squaring the term numbers and then following the pattern that they make. Which in this case is 1, 2, 3, 4, 5, ...
Eg) 1^2=1, 2^2=4, 3^2=9, 4^2=16... so 5^2=25
5, 8, 11, 14, ... The answer is 17. We get this by adding 3 to the previous number.
Eg) 5+3=8, 8+3=11, 11+3=14... so 14+3=17
2, 6, 18, .... The answer is 54. We get this answer by multiplying the previous number by 3.
Eg) 2x3=6, 6x3=18... so 18x3=54
The second Sequence is something called an Arithmetic Sequence.
Terms: t1 t2 t3 t4
Sequences: 5 5+(3) 5+2(3) 5+3(3)
How would you find the 100th term?
To find the 100th term we have to follow the rules that we just discussed;
t100= 5+99(3)
t100=302
Arithmetic Sequences:
An Arithmetic Sequence is a sequence like: 6, 10, 14, 18 ... in which each term can be calculated by adding a consistent value to the preceding term. The value is called the common difference, d.
There are two definitions/equations that you need to know for this type of Sequence:
*Recursive definition: tn=tn-1+d (you must know the previous term to find the next term)
*Explicit definition: tn=t1+(n-1)(d) Where:
d is the common difference
t1 is the value of the first term
n is the number of terms
tn is the value of the nth term
Example) Given the Arithmetic Sequence: 5, 13, 21, ...
A) Write a function to generate the Sequence:
tn=t1+(n-1)(d)
tn=5+(n-1)(8)
tn=5+8n-8
tn=8n-3
B) Find the value of the 18th term.
t18=8(18)-3
t18=141
Example 2) Given that an Arithmetic Sequence has t5=6 and t8=27, find term 2.
first of all, we know that there is always a difference between terms. So if we have term 5 and term 8, there will be three differences between them. We can
represent this as 3d.
So...
6+3d=27
3d=21
d=7
There is also a three number space between term 5 and term 2.
Therefore...
Term 2=6-3d
6-21=-15
t2=-15
Hopefully this helps!!!
Hopefully everyone was happy with their test results, had a great weekend, and finished the assignment that was due. Also, Happy belated "name day" to Mr. P's mom!
Oh yeah! We also stared learning a new unit.
Geometric Sequences & Series:
Lets start with the basics;
Example: Find the next term in each sequence. Describe how you got them.
1, 4, 9, 16, .... The answer is 25. We get this by squaring the term numbers and then following the pattern that they make. Which in this case is 1, 2, 3, 4, 5, ...
Eg) 1^2=1, 2^2=4, 3^2=9, 4^2=16... so 5^2=25
5, 8, 11, 14, ... The answer is 17. We get this by adding 3 to the previous number.
Eg) 5+3=8, 8+3=11, 11+3=14... so 14+3=17
2, 6, 18, .... The answer is 54. We get this answer by multiplying the previous number by 3.
Eg) 2x3=6, 6x3=18... so 18x3=54
The second Sequence is something called an Arithmetic Sequence.
Terms: t1 t2 t3 t4
Sequences: 5 5+(3) 5+2(3) 5+3(3)
How would you find the 100th term?
To find the 100th term we have to follow the rules that we just discussed;
t100= 5+99(3)
t100=302
Arithmetic Sequences:
An Arithmetic Sequence is a sequence like: 6, 10, 14, 18 ... in which each term can be calculated by adding a consistent value to the preceding term. The value is called the common difference, d.
There are two definitions/equations that you need to know for this type of Sequence:
*Recursive definition: tn=tn-1+d (you must know the previous term to find the next term)
*Explicit definition: tn=t1+(n-1)(d) Where:
d is the common difference
t1 is the value of the first term
n is the number of terms
tn is the value of the nth term
Example) Given the Arithmetic Sequence: 5, 13, 21, ...
A) Write a function to generate the Sequence:
tn=t1+(n-1)(d)
tn=5+(n-1)(8)
tn=5+8n-8
tn=8n-3
B) Find the value of the 18th term.
t18=8(18)-3
t18=141
Example 2) Given that an Arithmetic Sequence has t5=6 and t8=27, find term 2.
first of all, we know that there is always a difference between terms. So if we have term 5 and term 8, there will be three differences between them. We can
represent this as 3d.
So...
6+3d=27
3d=21
d=7
There is also a three number space between term 5 and term 2.
Therefore...
Term 2=6-3d
6-21=-15
t2=-15
Hopefully this helps!!!
Tuesday, January 10, 2012
Quadratic Inequalities in Two Variables
hey guys, this is ashley, i was supposed to blog yesterday but forgot so here goes...
Quadratic inequalities in two variables can be expressed in one of the four forms:
y < ax² + bx + c when either of these to inequalities is used you will graph
y < ax² + bx + c with a dotted line...
y ≤ ax ² + bx +c with either of these two inequalities, you will graph with
y ≥ ax² + bx + c a solid line
where a, b, and c are real numbers and a does not equal to zero
We know: because we have < in our inequality, we know that we are going to graph with a dotted line. that the x value serves as the axis of symmetry, and that the parabola will open downwards.
and that is how you would go about solving this quadratic inequality with two variables.
Quadratic inequalities in two variables can be expressed in one of the four forms:
y < ax² + bx + c when either of these to inequalities is used you will graph
y < ax² + bx + c with a dotted line...
y ≤ ax ² + bx +c with either of these two inequalities, you will graph with
y ≥ ax² + bx + c a solid line
where a, b, and c are real numbers and a does not equal to zero
- A quadratic inequality in two variables represents a region on a cartesian plain with a parabola as the boudary. The graph of a quadratic inequality is the set of points (x,y) which are solutions to the inequality.
Example 1.
graph y< -2 (x - 3)² + 1. Is the point (2, -4) an inequality?We know: because we have < in our inequality, we know that we are going to graph with a dotted line. that the x value serves as the axis of symmetry, and that the parabola will open downwards.
- start with finding the vertex which is at (3, 1)
2. pick a point such as x=4. we have to find the y value and we will go about doing this by plugging the x value into the equation. y< -2 (x - 3)² + 1..... y< -2 (4-3)² +1 ..... y=-2(1)² +1 ...... y= -2 +1 ..... y=-1 Now we can reflect this point over the access of symmetry (2, -1)
3. find which area to shade. to do this you need two test points, one inside the parabola, and one outside the parabloa. you can pick any points, we will use (3, -3) inside, and (0,0) for the outside points. now just solve and find which is true and false. we will start with the inside points: y< -2 (x-3)² +1 .... -3 < -2(3-3)² +1 ... -3 < -2(0) +1 .... -3 < 1 which is true which means we will shade the inside of the parabola. to further prove this we can solve for the outside points and see if they are false: y<-2(x-3)+1 ... 0 < -2(0-3)²+1 ... 0 < -2 (9) +1... 0<-18+1 ... 0<-17 which is false.
and that is how you would go about solving this quadratic inequality with two variables.
Monday, December 19, 2011
Quadratic- Quadratic Systems
Hey guys, it's Jhelene again scribing for your post today! So it's early Monday morning after the elimination quiz we had, and we learned the second part to the Systems of Non-Linear Equations.
So you might be thinking "What the heck is a quadratic-quadratic system?" Well it is basically is a system of quadratic relations or what we learned as conics; and it may have 0, 1, 2, 3, or 4 real number solutions.
So you might be thinking "What the heck is a quadratic-quadratic system?" Well it is basically is a system of quadratic relations or what we learned as conics; and it may have 0, 1, 2, 3, or 4 real number solutions.
no real number solutions
one real number solution
two real number solutions
one real number solution
two real number solutions
three real number solutions
four real number solutions
While we didn't have any assignment besides Friday's worksheet, we did the examples on the booklet like this one:
So we start off with these two equations waiting to be solved..
Seeing as how both x and y values are right, we therefore say that the answer for these equations is (-4, 0).
I hope this post helped you a bit!
PS. Sorry for the crooked pictures up top, my scanner was messed LOL
PPS. 6 days before Christmas!!!! Yay okay bye haha.
While we didn't have any assignment besides Friday's worksheet, we did the examples on the booklet like this one:
So we start off with these two equations waiting to be solved..
Then I used one of the equations, and solved for y. In this case there were two possible y answers, so I plugged both into another equation. However, one of my answers did not surpass the equation and had to reject it. 
To make sure we got the right answers, plug in your x and y values into each equation and solve.
To make sure we got the right answers, plug in your x and y values into each equation and solve.
Seeing as how both x and y values are right, we therefore say that the answer for these equations is (-4, 0).
I hope this post helped you a bit!
PS. Sorry for the crooked pictures up top, my scanner was messed LOL
PPS. 6 days before Christmas!!!! Yay okay bye haha.
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