Monday, March 28, 2011
Nathan's blog
This week in advanced math, we basically did a bunch of aleks. Monday we had SAT's and everyone else went on a field trip. Then the rest of the week we started on chapter 6, and on thursday and friday, we did aleks with the sub. Chapter 6 had to deal with equations of circles and ellipses, and stuff like that. I checked out on wednesday and missed a little bit of the lesson but i should be able to pick up on it pretty fast. That is all for this blog.
Sunday, March 27, 2011
mary'ss blog
this. week. was. crazy. We had to sell candy and get ready for state, and we did aleks, alot of them, and then we couldn't do work because brob left us. haha. But dang, all them aleks really helped during everything i did at state, especially the game descartes. i swear i wouldn't have known half the stuff i knew without it. i really like math club, it's really fun and worth it, i can't even imagine not being in it at this point:)) this question came up on a 10 point card in how many zeros does 127! end! ahahahah!!!!!!!!!!1
Nicala's Post
this week we did not really get to do anything. because of the sat and the field trip and then thing with mu alpha theta. so we mainly did aleks the whole week. Which most of them i did not know what i am doing so i get put out or i fail my assestment and have to do the same things over and over. okay so i am going to back to chapter 13 section 1. In this chapter they are talking about sequences. A sequences is a set of number. There are two different type of sequences we learn about in this section, arithmetic and geometric. I am only talking about arithmetic because i am sleeping and ready to go to bed lol :) Arithmetic squence is when you add the same number to every term. The only formula you need to know for this problem is tn=t1+(n-1)d. d is the number that you are adding to each term. Example Problem Find the first four terms of this equation 7n+2. in the problem you always replace n with number term you are on. t1=7(1)+2=9 t2=7(2)+2=16 t3=7(3)+2=23 t4=7(4)+2=30 So the first four terms in this equation is 9,16,23,30
Lawrence's blog
this week was really crazy. we couldnt really do much in the class because of the SAT and people that were in mu alpha theta. so really all we did was Aleks. i didnt know how to do any of the ones that we had to do. i attempted even copied the problems and tried workign them step for step and still couldnt get them. so i guess im screwed till it makes me take a dumb 30 question test. if there is anyone who is below me and may need my help i can help you with it and hopefully for some odd reason someone will know how to do the stuff im on
Friday, March 25, 2011
Another One Of Taylor's Blogs
The following is part one of my notes on lesson 6.3.
In this lesson we will be doing the easier parts of finding the parts of the equation.
The one we will be doing are in this form
x^2/a+y^2/b=1
We will by finding the minor and major axis and the length of the major and minor axis.
Equations
Length of major 2 square root of lager denom.
Length of minor 2 square root of smaller denom.
Lets do some example problems
Find the minor and major axis and the length of the major and minor axis of x^2/9+y^2/25=1
First we will identify the major and minor axis. In this case the major axis is y and the minor axis is x.
2 square root of 25= 2 x 5= 10 Then we simply plug the major axis’s denominator into this equation and solve it to find the length of the major axis.
2 square root of 9= 2 x 3= 6 Then we simply plug the minor axis’s denominator into this equation and solve it to find the length of the minor axis.
And your answers will be major axis= y, minor axis= x, length of the major axis= 10, length of the minor axis= 6. Lets do one more!
Find the minor and major axis and the length of the major and minor axis of x^2/81+y^2/36=1
First we will identify the major and minor axis. In this case the major axis is x and the minor axis is y.
2 square root of 81= 2 x 9= 18 Then we simply plug the major axis’s denominator into this equation and solve it to find the length of the major axis.
2 square root of 36= 2 x 6= 12 Then we simply plug the minor axis’s denominator into this equation and solve it to find the length of the minor axis.
And your answers will be major axis= x, minor axis= y, length of the major axis= 18, length of the minor axis= 12.
THAT IS ALL FOR THIS PART OF MY LESSON 6.3 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
In this lesson we will be doing the easier parts of finding the parts of the equation.
The one we will be doing are in this form
x^2/a+y^2/b=1
We will by finding the minor and major axis and the length of the major and minor axis.
Equations
Length of major 2 square root of lager denom.
Length of minor 2 square root of smaller denom.
Lets do some example problems
Find the minor and major axis and the length of the major and minor axis of x^2/9+y^2/25=1
First we will identify the major and minor axis. In this case the major axis is y and the minor axis is x.
2 square root of 25= 2 x 5= 10 Then we simply plug the major axis’s denominator into this equation and solve it to find the length of the major axis.
2 square root of 9= 2 x 3= 6 Then we simply plug the minor axis’s denominator into this equation and solve it to find the length of the minor axis.
And your answers will be major axis= y, minor axis= x, length of the major axis= 10, length of the minor axis= 6. Lets do one more!
Find the minor and major axis and the length of the major and minor axis of x^2/81+y^2/36=1
First we will identify the major and minor axis. In this case the major axis is x and the minor axis is y.
2 square root of 81= 2 x 9= 18 Then we simply plug the major axis’s denominator into this equation and solve it to find the length of the major axis.
2 square root of 36= 2 x 6= 12 Then we simply plug the minor axis’s denominator into this equation and solve it to find the length of the minor axis.
And your answers will be major axis= x, minor axis= y, length of the major axis= 18, length of the minor axis= 12.
THAT IS ALL FOR THIS PART OF MY LESSON 6.3 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
Monday, March 21, 2011
Charlie.
Ok, so we didn't have any aleks due this week.
BUT, we did do stuff for math this week.
we learned chapter 6.
one thing we did was the eclipses & stuff like that.
the formula is: x^2/a + y^2/b = 1
to do this we have to find the major axis, the minor axis, the major length, the minor length, the vertex, other, and the focus.
EXAMPLE:
x^2/4 + y^2/16 = 1
Major --> y
Minor --> x
Major Length --> 8
Minor Length --> 4
Vertex --> (0, 4) (0, -4)
Other --> (2, 0) (-2, 0)
Focus --> (0, 2squarerootof3) (0, -2squarerootof3)
BUT, we did do stuff for math this week.
we learned chapter 6.
one thing we did was the eclipses & stuff like that.
the formula is: x^2/a + y^2/b = 1
to do this we have to find the major axis, the minor axis, the major length, the minor length, the vertex, other, and the focus.
EXAMPLE:
x^2/4 + y^2/16 = 1
Major --> y
Minor --> x
Major Length --> 8
Minor Length --> 4
Vertex --> (0, 4) (0, -4)
Other --> (2, 0) (-2, 0)
Focus --> (0, 2squarerootof3) (0, -2squarerootof3)
Sunday, March 20, 2011
Feroz's Blog
Haven't done one of these in awhile. Anyway, I'm gonna cover Chapter 6.
Chapter 6 deals with circles, parabolas, hyperboles and ellipses and stuff.
ex. x^2/16 + y^2/25 = 1
1. Major axis = variable with the greater denom = y
2. Minor axis = variable with the smaller denom = x
3. Length of Major = 2 x sqrt(major) = 10
4. Length of Minor = 2 x sqrt(minor) = 8
5. Vertex = since y is the major = (0,5), (0,-5)
6. Other int. = same thing with the minor = (4,0), (-4,0)
7. Focus = sqrt(major - minor) = sqrt(2) = (0, sqrt(2)), (0, -sqrt(2))
8. Sketch the graph
There you go.
Nathan's blog
This week, we started on chapter 6. It has to deal with circles, ellipses, and hyperbolas. So far it is pretty easy, cause the formulas and equations don't differ that much. Each one has a seven step process, and this basically helps you find all of the parts of the shape, and helps with plugging into the calculator, so that you can sketch the shape. Here's an example of finding the center and radius of a circle.
Ex. (x-3)^2 + (y+7)^2 = 19
C: (3,-7)
r= √19
Find all the parts of: x^2/4 + y^2/25 = 1
1. Major axis: y
2. Minor axis: x
3. Length of major: 2√25=10
4. Length of minor: 2√4=4
5. Vertex: (0,5) (0,-5)
6. Other intercept: (2,0) (-2,0)
7. Focus: 4=25-f^2 -21=-f^2 f=√21 (0,√21) (0,-√21)
That's about it for this review. The only difference in the other formulas, is that you have to find aymptotes. Otherwise, this chapter is very easy so far.
Ex. (x-3)^2 + (y+7)^2 = 19
C: (3,-7)
r= √19
Find all the parts of: x^2/4 + y^2/25 = 1
1. Major axis: y
2. Minor axis: x
3. Length of major: 2√25=10
4. Length of minor: 2√4=4
5. Vertex: (0,5) (0,-5)
6. Other intercept: (2,0) (-2,0)
7. Focus: 4=25-f^2 -21=-f^2 f=√21 (0,√21) (0,-√21)
That's about it for this review. The only difference in the other formulas, is that you have to find aymptotes. Otherwise, this chapter is very easy so far.
BLLOOGGG!
i love getting grades for blogs:)
thiiisss weeeeeeeeeek.
We are learning about chapter 6 about finding all the parts of a hyperbola and an ellipse.
there isn't really much to explain about it so i'll just walk you through the example.
Ex. 1 which is an ellipse, and you can tell the difference between and ellipse and a hyperbola by looking to see if there is a negative.
x^2/36 + y^2/49 =1 and to be in standard form, it will always have to equal one.
so to start it off
Major axis-y-because the bigger number is under the 7
minor axis-x- by default
vertex-(0,7),(0,-7) because of the major axis which they follow
focus- focus^2=bigger plus smaller=(0,plus or minus-square root of 85)-still using the place of the major axis
other intercept( which you use to sketch a graph)-(6,0)(-6,0) you use the square root of the minor axis in that place.
length of major axis-14
length of minor-12, you can get this by doubling the square root of the number under the x or y.
and the process is really close for a hyperbola, but there is the negative to deal with and you have to find asyptotes. but its really easy, everybody should really try really hard in this last semester and finish out strong and not have to worry about next year.::))
Nicala's Post
In class we are working on chapter six we did three sections already. the sections we worked on are circles, ellipses, and hyperbolas. Ellipses and hyperbolas are basically the same expect the change in their formulas and hyperbolas have asymptotes and ellipses do not asymptotes. Ellipses have a oval like shape. The formula for its
x^2/a^2+y^2/b^2=1
The formula for an hyperbola is the same except the y or x is negative and that changes the formula. When your working with ellipses you are looking for the major, minor, length of major, length of minor, vertex, other intercepts, and the focus and depending on the instructions you graph.
example problem
x^2/4+y^2/16=1
1.major-y because 16 is bigger than 4
2.minor is x because 4 is smaller than 16.
3. length of major: 2 the square root of 16= 8
4. length of minor: 2 the square root of 4=4
5.vertex is the square to the major and put the answer in points= (0,4) &(0,-4)
6. the other intercept is the square root of the minor and put the answer in points= (2,0) &(-2,0)
7.focus is the both denominators equal to f squared. 4-16=f^2 and now solve like normal and it gives u an answer of f=2 the square root of 3 and you put your final in points. (0, 2 the square root of 3) & (0,-2 the square root of 3)
8. sketching the graph is option
bloggggg
This weekk, we started learning chapter 6. this chapter is pretty much about graphs like hyperboles, ellipes, and circles.
Circle
equation: (x-h)^2+(y-k)^2=r^2
where (h,k) is center
To put an equation into the standard form of a circle, you have to complete the square.
Ellipse
equation: x^2/a + y^2/b=1
major axis: variable with larger denomenator
minor axis: varibale with smaller denomenator
length of major: 2(squarerootof larger)
length of minor: 2(squarerootof smaller)
vertex: (squarerootof larger) if x is major: ( , 0) (- , 0)
if y is major: (0, ) (0,- )
other int. : (squarerootof smaller) ( , )
focus: smaller=larger-focus^2
Hyperbole
This is pretty much the same as the ellipse, except you also have to deal with negative numbers and asymptotes. also the focus is different.
focus: focus^2=larger+smaller
asymptote: y=+-b/a x if x is major
y=+-a/b x if y is major
Circle
equation: (x-h)^2+(y-k)^2=r^2
where (h,k) is center
To put an equation into the standard form of a circle, you have to complete the square.
Ellipse
equation: x^2/a + y^2/b=1
major axis: variable with larger denomenator
minor axis: varibale with smaller denomenator
length of major: 2(squarerootof larger)
length of minor: 2(squarerootof smaller)
vertex: (squarerootof larger) if x is major: ( , 0) (- , 0)
if y is major: (0, ) (0,- )
other int. : (squarerootof smaller) ( , )
focus: smaller=larger-focus^2
Hyperbole
This is pretty much the same as the ellipse, except you also have to deal with negative numbers and asymptotes. also the focus is different.
focus: focus^2=larger+smaller
asymptote: y=+-b/a x if x is major
y=+-a/b x if y is major
Friday, March 18, 2011
Taylor's Mystery Number Blog (I GIVE UP ON NUMBERING IT T-T!!)
The following is part one of my notes on lesson 6.2.
In this lesson we will be learning a few new things about circles.
You will need these equations in order to do this properly:
Distance formula= square root (y2-y1)^2+(x2+x1)^2= radius
The 2 and ones are small and are at the bottom to symbolize that they are from either the first and second point . One of these points must be the center while the other must be on the circle.
Equation of a circle (x-h)^2+(y-k)^2=r^2 the (h,k) is the center of the circle.
Now lets practice how to find the radius and the center of the circle from the equation of the circle. Which is also the easiest and simplest part to do for this chapter.
Find the center and the radius of (x-9)^2+(y-8)^2=16.
From the equation of the circle we can see that the place where 9 and 8 are is replacing the h and the k so that means that the center of the circle is (9,8).
We also know that the radius is raised to the second before it can be considered part of the equation so we just need to find the square root of 16 in order to find the radius. So that makes the radius 4.
Find the center and the radius of (x+5)^2+(y+11)^2=15.
From the equation of the circle we can see that the place where 5 and 11 are is replacing the h and the k but since they are not negative we have to make them negative when we write it in coordinate form. So the center of the circle is (-5,-11)
We also know that the radius is raised to the second before it can be considered part of the equation so we just need to find the square root of 15 in order to find the radius but since we cannot we will simply leave it with the square root as the answer. So that makes the radius square root of 15.
THAT IS ALL FOR THIS PART OF MY LESSON 6.2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
In this lesson we will be learning a few new things about circles.
You will need these equations in order to do this properly:
Distance formula= square root (y2-y1)^2+(x2+x1)^2= radius
The 2 and ones are small and are at the bottom to symbolize that they are from either the first and second point . One of these points must be the center while the other must be on the circle.
Equation of a circle (x-h)^2+(y-k)^2=r^2 the (h,k) is the center of the circle.
Now lets practice how to find the radius and the center of the circle from the equation of the circle. Which is also the easiest and simplest part to do for this chapter.
Find the center and the radius of (x-9)^2+(y-8)^2=16.
From the equation of the circle we can see that the place where 9 and 8 are is replacing the h and the k so that means that the center of the circle is (9,8).
We also know that the radius is raised to the second before it can be considered part of the equation so we just need to find the square root of 16 in order to find the radius. So that makes the radius 4.
Find the center and the radius of (x+5)^2+(y+11)^2=15.
From the equation of the circle we can see that the place where 5 and 11 are is replacing the h and the k but since they are not negative we have to make them negative when we write it in coordinate form. So the center of the circle is (-5,-11)
We also know that the radius is raised to the second before it can be considered part of the equation so we just need to find the square root of 15 in order to find the radius but since we cannot we will simply leave it with the square root as the answer. So that makes the radius square root of 15.
THAT IS ALL FOR THIS PART OF MY LESSON 6.2 NOTES !! SO UNTIL NEXT TIME JA NE!!!! (ja ne is goodbye in Japanese)
Tuesday, March 15, 2011
Charlie
this week we had the holidays..
umm, i did 39 of my 50 bonus point modules.
my computer was broken until wednesday so i did it thursday and friday.
& then a couple more on sunday. i hope not dong this doesn't make me fail. but whateva.
i don't remember what we were learning before we went into the holidays.
but on aleks i didn't know how to do the blue, purple, or yellow stuff.
so i went do some of the trig. stuff & what i had unlocked for the matrices thing.
but i still didn't know how to do anything else to get to my 50 modules.
so that isn't fair, JUS SAYin.
it's hot in this class.
umm, i did 39 of my 50 bonus point modules.
my computer was broken until wednesday so i did it thursday and friday.
& then a couple more on sunday. i hope not dong this doesn't make me fail. but whateva.
i don't remember what we were learning before we went into the holidays.
but on aleks i didn't know how to do the blue, purple, or yellow stuff.
so i went do some of the trig. stuff & what i had unlocked for the matrices thing.
but i still didn't know how to do anything else to get to my 50 modules.
so that isn't fair, JUS SAYin.
it's hot in this class.
Sunday, March 13, 2011
Nathan's blog
Yeah, so i didnt do my aleks and now I'm mad, so i might as well get some points for this. I'm probably just going to do a basic review of stuff we learned in class, and stuff in aleks.
So, I'm going to start off by reviewing matrices, which was very easy.
[2 4] [8 3] [10 7]
[7 5] + [9 12] = [16 17]
Now on to subtracting matrices.
[18 6] [5 7] [13 -1]
[15 3] - [14 9] = [ 1 -6]
Now, i will review how to find the slope of a line.
(4,3) (7,2)
2-4/7-3
The slope of the line is: -1/2
The slope of a line is 4/3. Find the slope of a parallel and perpendicular line, using this slope.
For the slope of a parallel line, it is the same as the slope of the original line. So the slope of this parallel line is: 4/3
When finding the slope of a perpendicular line, you take the negative reciporcal of the original slope. So the slope of this perpendicular line is: -3/4
That's it for this review.
So, I'm going to start off by reviewing matrices, which was very easy.
[2 4] [8 3] [10 7]
[7 5] + [9 12] = [16 17]
Now on to subtracting matrices.
[18 6] [5 7] [13 -1]
[15 3] - [14 9] = [ 1 -6]
Now, i will review how to find the slope of a line.
(4,3) (7,2)
2-4/7-3
The slope of the line is: -1/2
The slope of a line is 4/3. Find the slope of a parallel and perpendicular line, using this slope.
For the slope of a parallel line, it is the same as the slope of the original line. So the slope of this parallel line is: 4/3
When finding the slope of a perpendicular line, you take the negative reciporcal of the original slope. So the slope of this perpendicular line is: -3/4
That's it for this review.
Friday, March 11, 2011
Taylor's 14th(I think XD) Blog Review
REVIEW TIME!!!!
This is the review on my notes from Ch. 13.2.
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=4 tn=8tn-1+7
t1=8 (4)+7= 39 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 8 (39) + 7 = 319
t3= 8 (319) + 7= 2559
t4= 8 (2559) + 7 = 20479
t5= 8 (20479) + 7 = 163839
So your answers will be 2559, 20479, and 163839.
Now let's try finding the recursive definition for 8, 12, 16, 20...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 4 in between each number your answer or equation will be tn-1+4.
This is the review on the rest of my notes on Ch. 13.2. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
This is the review on my notes from Ch. 13.2.
A recursive means define in terms of what came before.
This is what you need to remember:
tn-1 previous term
tn-2 2 terms back
tn-3 ....
Ok let's do some examples.
Find the third, fourth, and fifth terms given. t1=4 tn=8tn-1+7
t1=8 (4)+7= 39 First replace tn-1 with the previous terms. You will keep doing this until you get all three answers. Remember to use your answers and not t1!!
t2= 8 (39) + 7 = 319
t3= 8 (319) + 7= 2559
t4= 8 (2559) + 7 = 20479
t5= 8 (20479) + 7 = 163839
So your answers will be 2559, 20479, and 163839.
Now let's try finding the recursive definition for 8, 12, 16, 20...
First, you need to figure out how much is in-between each number and how to get to the next number with the previous numbers help.
Since there is 4 in between each number your answer or equation will be tn-1+4.
This is the review on the rest of my notes on Ch. 13.2. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
Sunday, March 6, 2011
Nathan's blog
So, we just took our exams on chapter 5 which dealt with logs and interest, and a few more formulas. I thought I understood the information, but apparently not. I just finished the 14 aleks modules that we needed done by today, and then tommorrow i will start on the 50 modules for extra credit.
So, now i will just review different stuff throughout chapter 5.
Rule of 72
72/r%=how long it takes to double
How long will it take to double your money at a 12% interest?
72/12=6
We learned a formula for interest:
A(t)=Ao b^t/k
Ex. A bank advertises that if you open a savings account, you can double your money in twelve years. Find out how much money you will have after 7 years, if you invest $1,000 into your account.
A(t)=1000(2)^7/12=$1498.31
I'm very happy to get out of chapter 5, hopefully i will understood the information better in the 4th nine weeks, because that will probably be the toughest part of the year in math.
So, now i will just review different stuff throughout chapter 5.
Rule of 72
72/r%=how long it takes to double
How long will it take to double your money at a 12% interest?
72/12=6
We learned a formula for interest:
A(t)=Ao b^t/k
Ex. A bank advertises that if you open a savings account, you can double your money in twelve years. Find out how much money you will have after 7 years, if you invest $1,000 into your account.
A(t)=1000(2)^7/12=$1498.31
I'm very happy to get out of chapter 5, hopefully i will understood the information better in the 4th nine weeks, because that will probably be the toughest part of the year in math.
Nicala's Post
okay we had our exams and we learned the rest of chapter five which was about exponents which we learned in aleks and logs which we should have learned in algebra 2 but i did not understand it then and do not understand now. okay we took our exam thursday which i failed horrible and we started doing fourteen aleks friday and will finish them this weekend and tomorrow we start the fifty bonus aleks for the extra fifty points in class which i need. one of things i learned on aleks is how to do translation of a vector.
Example Problem
A vector with an initial point of (-6,-2) and terminal point of (-5,3) is translated so that its initial point is at the origin. Find its new terminal point.
To find the terminal point you have subtract them from each other.
you subtract -5 minus -6. since you are subtracting and the six is negative the six becomes positive which gives you 1 as your first new terminal point.
next you subtract 3 minus -2 since you are subtracting and the two is the two becomes positive
which gives you 5 as your second new terminal point.
the answer is (1,5)
Example Problem
A vector with an initial point of (-6,-2) and terminal point of (-5,3) is translated so that its initial point is at the origin. Find its new terminal point.
To find the terminal point you have subtract them from each other.
you subtract -5 minus -6. since you are subtracting and the six is negative the six becomes positive which gives you 1 as your first new terminal point.
next you subtract 3 minus -2 since you are subtracting and the two is the two becomes positive
which gives you 5 as your second new terminal point.
the answer is (1,5)
Friday, March 4, 2011
Taylor's 14th(I think XD) Blog Review
REVIEW TIME!!!!
This is the review on the rest of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 3 x 2^ n-1
t1=3 x 2^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=3 x 2^ 2-1
t3=3 x 2^ 3-1
t4=3 x 2^ 4-1
t1= 3 Then you have your answers.
t2= 6
t3= 12
t4= 24
9,36,144,576.... find the formula for the nth term.
9,36,144,576.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 4.
tn= t1 x r ^n-1Then write out your equation.
tn= 9 x4 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
This is the review on the rest of my notes on Ch. 13.1. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
This is the review on the rest of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about geometric sequence which is when you multiply the same number to every term.
This is the formula:
tn= t1 x r ^n-1
r is the number that is being multiplied or divided. The n and one by the t are small to.
Ok let's do some examples.
Find the first 4 terms of this equation tn= 3 x 2^ n-1
t1=3 x 2^ 1-1 First replace n with 1,2,3,and 4 and solve the equation.
t2=3 x 2^ 2-1
t3=3 x 2^ 3-1
t4=3 x 2^ 4-1
t1= 3 Then you have your answers.
t2= 6
t3= 12
t4= 24
9,36,144,576.... find the formula for the nth term.
9,36,144,576.... First you need to figure out what is being multiplied or divided from each number. In this case it is multiplying by 4.
tn= t1 x r ^n-1Then write out your equation.
tn= 9 x4 ^n-1 Then fill out the parts that you know and since you cannot do anything else to this equation this is your answer.
This is the review on the rest of my notes on Ch. 13.1. So SEE YALL!!! JA NE!!!(Japanese word for goodbye)
Charlie.
This week we had exams.
But anyways, in aleks i ended up having to take the stupid assessment thingy & got knocked down to like 70 in my blue thingy..
It's friday, in first hour.
& i think today in math class we'll be coming to the lab or A1 to do 7 alek modules.
but not the library because the book fair is this week, so yea.
then, on monday we could start on the extra credit thingy.
extra credit this nine weeks is 50 aleks modules.
which, idk how i'm goin to do that because MY COMPUTER'S BROkEN!
but, i guess i'll figure something out because i would like a really good grade this time.
... This week we learned 2 (or 3?) new sections in chapter (i think) 5.
then we pretty much reviewed for the rest of the week and took our exam yesterday (thursday).
when i was doing the practice chapter test in the text book i knew what i was doing,
but when i actually took the exam; i was like "WhAt?!?"
so, (basically) i do believe that i FAILED(:
example of something:
log2~x = 4
2^4 = x
x = 16
But anyways, in aleks i ended up having to take the stupid assessment thingy & got knocked down to like 70 in my blue thingy..
It's friday, in first hour.
& i think today in math class we'll be coming to the lab or A1 to do 7 alek modules.
but not the library because the book fair is this week, so yea.
then, on monday we could start on the extra credit thingy.
extra credit this nine weeks is 50 aleks modules.
which, idk how i'm goin to do that because MY COMPUTER'S BROkEN!
but, i guess i'll figure something out because i would like a really good grade this time.
... This week we learned 2 (or 3?) new sections in chapter (i think) 5.
then we pretty much reviewed for the rest of the week and took our exam yesterday (thursday).
when i was doing the practice chapter test in the text book i knew what i was doing,
but when i actually took the exam; i was like "WhAt?!?"
so, (basically) i do believe that i FAILED(:
example of something:
log2~x = 4
2^4 = x
x = 16
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