Blah blah blah blue pie blah blah Aleks blah Chapter 5.
- Rule 72
ex. How long would it take to double your money at 36% interest?
72/36 = 2
- Rational Exponents
ex. x^1/2
= sqrt(x)
Covered two sections without complaining about exams or a bee in my kitchen. I think this went pretty well.
Sunday, February 27, 2011
Mary
|
A street that is 222 m long is covered in snow. City workers are using a snowplos to clear the street. A tire on the snowplow has to turn 37 times n traveling the length of the street. What is the diameter of the tire? crazy right?
| Here is the solution, it took me forever to get this module correct! it took me all class! First, we'll find the circumference C of the tire. Then, we'll use C to calculate the diameter d .
We are told that the tire turns
| ||||||||||||||||||||||||||||||||
Lawrence"s blog
this week we pretty much did chapter 12. we did all of the stuff she taught us on aleks so none of it is really new. it was more of a review than anything. All of chapter 12 seems like a review because of aleks wich is a good thing because we get to learn it and understand it and it doesnt seem like greek anymore. chapter 12 is also our exam i think too if i remember correctly. i wont put an examples cuz you can go on aleks and find them or you can look at the one's brob did but if you need help i will be glad to do so
Nathan's blog
So, this week all we did was review the stuff from chapter 5, which was basically an overview of our blue pie in aleks. This stuff was really easy, and this is the only thing the exam is on.
Section 5-2 was on rational exponents.
-The denominator tells you the root. The numerator is the exponent.
Ex. x^3/2= √x^3
In section 5-3, we learned a new formula.
-A(t)=Ao b^t/x
-Ao=starting amount
-b=Doubling, halfing, triplin, etc.
-t=time
-k=how long it takes to get b.
Ex. A bank advertises that if you open a savings account, you can double your money in 12 years. Find out how much money you will have after 7 years, if you invest $1,000.
Ao=1,000 b=2
t=7 k=12
A(t)=1,000(2)^7/12= $1498.31
Rule of 72
72/r%=how long it takes to double.
Ex. How long will it take to double your money at 12% interest?
72/12=6
Chapter 5 is pretty easy, and I should do good on the exam, because I understand most of it due to the fact that it is basically a review of what we've been doing.
Section 5-2 was on rational exponents.
-The denominator tells you the root. The numerator is the exponent.
Ex. x^3/2= √x^3
In section 5-3, we learned a new formula.
-A(t)=Ao b^t/x
-Ao=starting amount
-b=Doubling, halfing, triplin, etc.
-t=time
-k=how long it takes to get b.
Ex. A bank advertises that if you open a savings account, you can double your money in 12 years. Find out how much money you will have after 7 years, if you invest $1,000.
Ao=1,000 b=2
t=7 k=12
A(t)=1,000(2)^7/12= $1498.31
Rule of 72
72/r%=how long it takes to double.
Ex. How long will it take to double your money at 12% interest?
72/12=6
Chapter 5 is pretty easy, and I should do good on the exam, because I understand most of it due to the fact that it is basically a review of what we've been doing.
Nicala's Post
In class this we learned chapter five which basically the same stuff that we have been doing on the blue part of the pie on aleks. We also started working on the gold and purple parts on the pie which are really hard so I did some of the modules in the green part to make the 14 that i need.
One of things that i worked on aleks is was the lines with slope and a point on the line.
Example problem
A line passes through the point (-5,-3) and has a slope of 2. Write an equation of the line.
The form that you use in this problem is y=mx+b. You put the numbers in place of yr variables.You use -5 as yr x and -3 as yr y.
-3=-5(2)+b
-3=-10+b
you move the -10 to other side of the equal sign and it becomes +1o and you add +10 to -3
which gives me +7
7=b
now you put the slope,2, and yr b,2, back in the problem to get yr answer.
y=2x+7
One of things that i worked on aleks is was the lines with slope and a point on the line.
Example problem
A line passes through the point (-5,-3) and has a slope of 2. Write an equation of the line.
The form that you use in this problem is y=mx+b. You put the numbers in place of yr variables.You use -5 as yr x and -3 as yr y.
-3=-5(2)+b
-3=-10+b
you move the -10 to other side of the equal sign and it becomes +1o and you add +10 to -3
which gives me +7
7=b
now you put the slope,2, and yr b,2, back in the problem to get yr answer.
y=2x+7
Friday, February 25, 2011
Taylor's 13th(I think XD) Blog Review
REVIEW TIME!!!!
This is a review on some of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about arithmetic sequence which is when you add the same number to every term.
This is the formula:
tn= t1 +(n-1)d
d is the number that is being added or subtracted. The n and one by the t are small to.
Ok let's do some examples.
Find the first 5 terms of this equation tn= 8n+4
t1= 8(1)+4 First replace n with 1,2,3, and for and solve the equation.
t2=8(2)+4
t3=8(3)+4
t4=8(4)+4
t1= 12 Then you have your answers.
t2= 20
t3= 28
t4= 36
7,18,29,40.... find the formula for the nth term.
7,18,29,40.... First you need to figure out what is being added or subtracted from
each number. In this case it is adding 11.
tn=t1+(n-1)d Then write out your equation.
tn=7+(n-1)11 Then fill out the parts that you know2.
tn=7+11n-11 Then multiply 5 times n and -1.
tn= -4 + 11n Then you just add up your like terms and this is your answer.
This was the review on some of my notes from Ch. 13.1.
P.s. SOS THERE IS A BEE IN MY HOUSE!!! T-T HELP!! I CAN'T EVEN DO MY HOME WORK!! ALL OF MY STUFF IS IN THE KITCHEN AND IT IS IN THERE TOO!!! T-T
This is a review on some of my notes from Ch. 13.1.
A sequence is a set of numbers. Today we will learn a bit about arithmetic sequence which is when you add the same number to every term.
This is the formula:
tn= t1 +(n-1)d
d is the number that is being added or subtracted. The n and one by the t are small to.
Ok let's do some examples.
Find the first 5 terms of this equation tn= 8n+4
t1= 8(1)+4 First replace n with 1,2,3, and for and solve the equation.
t2=8(2)+4
t3=8(3)+4
t4=8(4)+4
t1= 12 Then you have your answers.
t2= 20
t3= 28
t4= 36
7,18,29,40.... find the formula for the nth term.
7,18,29,40.... First you need to figure out what is being added or subtracted from
each number. In this case it is adding 11.
tn=t1+(n-1)d Then write out your equation.
tn=7+(n-1)11 Then fill out the parts that you know2.
tn=7+11n-11 Then multiply 5 times n and -1.
tn= -4 + 11n Then you just add up your like terms and this is your answer.
This was the review on some of my notes from Ch. 13.1.
P.s. SOS THERE IS A BEE IN MY HOUSE!!! T-T HELP!! I CAN'T EVEN DO MY HOME WORK!! ALL OF MY STUFF IS IN THE KITCHEN AND IT IS IN THERE TOO!!! T-T
Sunday, February 20, 2011
Feroz's Blog
This week we learned more on Chapter 12. Woohoo.
One thing I remember clearly is finding the magnitude.
ex. u = <8,3>
Find [u] <-- That's supposed to be an absolute value sign but my laptop doesn't have them.
sqrt(8^2 + 3^2)
= sqrt(73)
And dot product, that's always fun.
ex. <1,3> x <7,4>
* Multiply across and add.
7 + 12 = 19
One thing I remember clearly is finding the magnitude.
ex. u = <8,3>
Find [u] <-- That's supposed to be an absolute value sign but my laptop doesn't have them.
sqrt(8^2 + 3^2)
= sqrt(73)
And dot product, that's always fun.
ex. <1,3> x <7,4>
* Multiply across and add.
7 + 12 = 19
mmmmaaaaarrrrrrryyyyyyyy
This week we learned about chapter 12 and did a billion aleks.
Here are some random notes that i remember because i'm pretty sure i aced that test.
the symbols for absolute value mean the magnitude, which is where you square each term, then take the square root of that number.
l<5,4>l
5^2+4^2=25+16=41
square root of 41 is the magnitude of your vector.
then there's dot product, which they indicate by putting a dot inbetween two vectors. this gives you a number, NOT A VECTOR. hahaha.
<8,0> . <1,2>
then you just multiply across and you may be tempted to believe the numbers multiply across and then stay as a vector, but no, you have to add them.
if the dot product of two vectors equals zero, then they are perpendicular. just thought you'd like to know.
answer 8
they could also ask you to find the cos of the angle between two vectors, but don't fret, its easy if you recognize the examples above.
its the dot product of the vectors over the magnitude of each vector multiplied. you can figure it out by yourselfffffffffff.
Here are some random notes that i remember because i'm pretty sure i aced that test.
the symbols for absolute value mean the magnitude, which is where you square each term, then take the square root of that number.
l<5,4>l
5^2+4^2=25+16=41
square root of 41 is the magnitude of your vector.
then there's dot product, which they indicate by putting a dot inbetween two vectors. this gives you a number, NOT A VECTOR. hahaha.
<8,0> . <1,2>
then you just multiply across and you may be tempted to believe the numbers multiply across and then stay as a vector, but no, you have to add them.
if the dot product of two vectors equals zero, then they are perpendicular. just thought you'd like to know.
answer 8
they could also ask you to find the cos of the angle between two vectors, but don't fret, its easy if you recognize the examples above.
its the dot product of the vectors over the magnitude of each vector multiplied. you can figure it out by yourselfffffffffff.
Dylan's blog-o'-blogs
Ok, so we learned about vectors. We learned about some matrices and other things too, but mostly vectors and stuff. We learned about dot multiplying...which I will be talking about.
Vector 1= (x1,y1), Vector 2= (x2,y2)
To find the dot product:
Vector1*vector2= x1(x2)+y1(y2)
If x=(4,3) & q=(0,6)
then
x*q=4(0)+3(6)
=0+18
=18
(This product will always be a number)
Vector 1= (x1,y1), Vector 2= (x2,y2)
To find the dot product:
Vector1*vector2= x1(x2)+y1(y2)
If x=(4,3) & q=(0,6)
then
x*q=4(0)+3(6)
=0+18
=18
(This product will always be a number)
Nathan's blog
So, we had a test on chapter 12 this past week, and i will just give you the notes and some examples from random sections.
Vectors have a magnitude and a direction.
To add vectors, you add head to tail. The triangle formed by the vectors give the resultant vector.
To add vectors algebraically, you add the components.
Scalar multiplication puts a scalar times each compound.
Parametric equations take only the x portion or y portion.
Dot product
-v1= (x1,y1) v2= (x2,y2)
-v1*v2= x1x2+y1y2
Dot product is a scalar product meaning it gives you a number.
Ex. If u=(3,6), v=(4,2) and w=(-12,-6). Find u*v and v*w. Show that u and v are perpendicular and v & w are parallel.
u*v=(3)(4)+(-6)(2)=0
v*w=(4)(-12)+(2)(-6)=-60
Since the dot product of u*v is 0, then they are perpendicular.
v=(4,2) w=(-12,-6)
v=2(2,1) w=-6(2,1)
V and W are parallel and multiples of each other.
That's it for chapter 12, im glad we're finished with it, because it was kinda hard.
Vectors have a magnitude and a direction.
To add vectors, you add head to tail. The triangle formed by the vectors give the resultant vector.
To add vectors algebraically, you add the components.
Scalar multiplication puts a scalar times each compound.
Parametric equations take only the x portion or y portion.
Dot product
-v1= (x1,y1) v2= (x2,y2)
-v1*v2= x1x2+y1y2
Dot product is a scalar product meaning it gives you a number.
Ex. If u=(3,6), v=(4,2) and w=(-12,-6). Find u*v and v*w. Show that u and v are perpendicular and v & w are parallel.
u*v=(3)(4)+(-6)(2)=0
v*w=(4)(-12)+(2)(-6)=-60
Since the dot product of u*v is 0, then they are perpendicular.
v=(4,2) w=(-12,-6)
v=2(2,1) w=-6(2,1)
V and W are parallel and multiples of each other.
That's it for chapter 12, im glad we're finished with it, because it was kinda hard.
Lawrence's blog
we really didnt do anything new this whole week. it was just review for our chapter test and try to finish the blue chart on the pie. the chapter test wasnt really all that hard to me, it was just alot of finding the equations, dot product and it was just alot of formulas we had to know for chapter 12. i think i got the four by four right and cramers rule i totally forgot. if anyone needs help i can help them with chapter 12 because i got a pretty good hang of it and thank god my friend let me use his laptop to do my blog
Friday, February 18, 2011
Nicala's Blog
The week we have been working on chapter twelve which is vectors and also had matrices like in chapter four but now the matrices are four by four. We also working on the aleks and it gave me an assestment quiz which i bombed and it took like twelve modules from me but its said i went up a couple point i do not know how i went up and fail the assestment. Lately on aleks i have been working on simplify radical expressions which is very so i want to do an example problem of that.
example problem
simplify the following problem
square root of 18x^5y^4z
okay simplify basically means breaking it down to its lowing forms
1. for 18 you can take the square root of nine and that lefts two up the square root
2. x^5 you can take 4 of the Xs because there is five so you put x^2 outside of the square and leave the other x in the square root
3.the same way for y^4 there are four ways so you can take them out and its not an out number so you can take all of it out
4. for the z there is only one of them that why there is no exponent next to it and since there isnt two or more you have leave
it in the square root
the answer is 3x^2y^2 with square root of 2xz
example problem
simplify the following problem
square root of 18x^5y^4z
okay simplify basically means breaking it down to its lowing forms
1. for 18 you can take the square root of nine and that lefts two up the square root
2. x^5 you can take 4 of the Xs because there is five so you put x^2 outside of the square and leave the other x in the square root
3.the same way for y^4 there are four ways so you can take them out and its not an out number so you can take all of it out
4. for the z there is only one of them that why there is no exponent next to it and since there isnt two or more you have leave
it in the square root
the answer is 3x^2y^2 with square root of 2xz
Taylor's 10th(I thinkXD) Blog Review!!
REVIEW TIME!!!
This is a review on the rest of my half of my notes on section 10.1. We will only be doing cos for this section.
The formula that you need to know for this part of the section is this:
cos (alpha+/- beta) = cosalphaxcosbeta -/+ sinalphasinbeta
Alpha and Beta can come from your trig chart. Remember that you need to add or subtract to get the angle that you are looking for.
First Equation
Find the exact value of cos 90 degrees.
cos ( 60+30)= cos 60 degrees x cos 30 degrees - sin 60 degrees x sin 30 degrees First, you need to figure out what angles in the trig chat equal 90 degrees and if you need to subtract or add to get to it. Since 60 degrees plus 30 degrees equals 90 degrees, you will use the subtraction formula for this equation. Then you simply fill out your formula with the information that you were given.
1/2 x square root of 3/2 - square root of 3/2 x ½ Then you find out the trig functions by using your trig chart.
square root of 3/4 - square root of 3/4 Then you multiply the 2 sides by themselves.
cos 120 degrees= 0 Finally you subtract them and you get this as your answer.
That is how you use the formula to find the exact value for cos.
Now lets learn how to simplify.
Simplify cos 45 degrees x cos 60 degrees + sin 45 degrees x sin 60 degrees.
cos 60 degrees x cos 45 degrees + sin 60 degrees x sin 45 degrees First, write out your equation.
cos ( 60 degrees- 45 degrees) Then, since you know that you’re using cos’s formula and subtraction thanks to the formula. You will put it like this.
cos 15 degrees Then you subtract the two degrees and you get this.
cos 15 degrees Since 15 degrees is on the trig chart you will just leave it as the answer to your question.
That is how you use the cos formula in these situations.
This is the review on my notes on the rest of section 10.1.
SO UNTIL NEXT WEEK JA NE!!!
This is a review on the rest of my half of my notes on section 10.1. We will only be doing cos for this section.
The formula that you need to know for this part of the section is this:
cos (alpha+/- beta) = cosalphaxcosbeta -/+ sinalphasinbeta
Alpha and Beta can come from your trig chart. Remember that you need to add or subtract to get the angle that you are looking for.
First Equation
Find the exact value of cos 90 degrees.
cos ( 60+30)= cos 60 degrees x cos 30 degrees - sin 60 degrees x sin 30 degrees First, you need to figure out what angles in the trig chat equal 90 degrees and if you need to subtract or add to get to it. Since 60 degrees plus 30 degrees equals 90 degrees, you will use the subtraction formula for this equation. Then you simply fill out your formula with the information that you were given.
1/2 x square root of 3/2 - square root of 3/2 x ½ Then you find out the trig functions by using your trig chart.
square root of 3/4 - square root of 3/4 Then you multiply the 2 sides by themselves.
cos 120 degrees= 0 Finally you subtract them and you get this as your answer.
That is how you use the formula to find the exact value for cos.
Now lets learn how to simplify.
Simplify cos 45 degrees x cos 60 degrees + sin 45 degrees x sin 60 degrees.
cos 60 degrees x cos 45 degrees + sin 60 degrees x sin 45 degrees First, write out your equation.
cos ( 60 degrees- 45 degrees) Then, since you know that you’re using cos’s formula and subtraction thanks to the formula. You will put it like this.
cos 15 degrees Then you subtract the two degrees and you get this.
cos 15 degrees Since 15 degrees is on the trig chart you will just leave it as the answer to your question.
That is how you use the cos formula in these situations.
This is the review on my notes on the rest of section 10.1.
SO UNTIL NEXT WEEK JA NE!!!
Charlie's.
this week we reviewed for our test.
& yesturday we took our chapter test (it was half mulitple choice & half free response).
i think i got a good grade on it, idk though.
i knew how to do the cramer's rule thing & the 4 by 4 so i think that's the most important thing with the most points..
We also did 14 aleks' modules & have to do 10 modules over the weekend.
(which is a total of 24)
aleks is probably the only reason i'm passing right now.
(i got a 4.0 on my progess report:])
i don't know what else to write on here.
& i have absolutely no idea what example problems to do
because we haven't done new stuff.
K? THANKS. BYE(:
& yesturday we took our chapter test (it was half mulitple choice & half free response).
i think i got a good grade on it, idk though.
i knew how to do the cramer's rule thing & the 4 by 4 so i think that's the most important thing with the most points..
We also did 14 aleks' modules & have to do 10 modules over the weekend.
(which is a total of 24)
aleks is probably the only reason i'm passing right now.
(i got a 4.0 on my progess report:])
i don't know what else to write on here.
& i have absolutely no idea what example problems to do
because we haven't done new stuff.
K? THANKS. BYE(:
Monday, February 14, 2011
Dylan's Blog
I wasn't at school for three days. Blah. I did get the last few notes though.
Determinate's for 3x3 matrices:
1. pick a row r1->most 0s or 1s
2. delete row, delete column
3. repeat for other columns
ex.
|0 3 4|
|2 9 1|
|6 2 1|
this would become
|0 3 4|
|2 9 1|
|6 2 1| x=deleted
this leaves:
|2 9|
|6 2|
Now for the others:
|2 1|
|6 1|
&
|9 1|
|2 1|
Next you take the first number of the row and multiply by the matrix by the number:
|2 9|
|6 2|
|0 3 4|
|2 9 1|
|6 2 1|
4 would be the number used to multiply the matrix
4|2 9|
|6 2|
=
-200 (use ad-bc to find the matrix)
repeat for others, and you should get
0+24-200=-176
Determinate's for 3x3 matrices:
1. pick a row r1->most 0s or 1s
2. delete row, delete column
3. repeat for other columns
ex.
|0 3 4|
|2 9 1|
|6 2 1|
this would become
|0 3 4|
|2 9 1|
|6 2 1| x=deleted
this leaves:
|2 9|
|6 2|
Now for the others:
|2 1|
|6 1|
&
|9 1|
|2 1|
Next you take the first number of the row and multiply by the matrix by the number:
|2 9|
|6 2|
|0 3 4|
|2 9 1|
|6 2 1|
4 would be the number used to multiply the matrix
4|2 9|
|6 2|
=
-200 (use ad-bc to find the matrix)
repeat for others, and you should get
0+24-200=-176
Sunday, February 13, 2011
Feroz's Blog
This week was a blur. I remember finding determinates but that takes a long time to do, especially on a computer. 3x3 might not, but 4x4 just kick my ass. So, since everyone else review things they thought were difficult on Aleks, I'll review something I find difficult as well.
Finding Domain and Range:
Polynomials - Domain (-infinity, infinity)
Range: (-infinity, infinity) if odd
Square Roots - Domain:
1. Set inside equal to 0, solve for x
2. Set up a number line
3. Plug in the numbers, use the non-negative intervals
Range: [Vertical shift, infinity)
Fractions - Domain:
1. Set bottom equal to 0, solve for x
2. Set up intervals
Range:
1. Take limit as x -> infinity
2. Set up intervals
Absolute value - Domain: (-infinity, infinity)
Range: [shift, infinity) or (-infinity, shift)
Ex. x^3 + 2x - 3
x - 3 = 0
x = 0
Domain: (-infinity, 3) u (3, infinity)
Range: [1, infinity)
Done.
Finding Domain and Range:
Polynomials - Domain (-infinity, infinity)
Range: (-infinity, infinity) if odd
Square Roots - Domain:
1. Set inside equal to 0, solve for x
2. Set up a number line
3. Plug in the numbers, use the non-negative intervals
Range: [Vertical shift, infinity)
Fractions - Domain:
1. Set bottom equal to 0, solve for x
2. Set up intervals
Range:
1. Take limit as x -> infinity
2. Set up intervals
Absolute value - Domain: (-infinity, infinity)
Range: [shift, infinity) or (-infinity, shift)
Ex. x^3 + 2x - 3
x - 3 = 0
x = 0
Domain: (-infinity, 3) u (3, infinity)
Range: [1, infinity)
Done.
blog...:(
so aleks thinks i'm stupid and wants to know if i can add fractions with common denominators. its SUPER difficult...(sarcasm) so here are some steps and an example for you slow folk. and i actually think i got a negative wrong on the assesment and had to do this module again! its really simple i promise. i would go over what ya'll did friday but i wasn't here and someone should really give me the notes tomorrow.
Problem: (4v-3w)/6v-(9v+5w)/6v
Each fraction has the same denominator, namely
So we subtract the numerators and keep the denominator the same to get
(4v-3w)-(9v+5w)
----------------------
6v
Distributing the negative sign over the parentheses, we have
4v-3w-9v-5w
-----------------
6v
Combining like terms, we get
-5v-8w
---------
6v
This fraction cannot be simplified so that's the answer. have fun.
malorie's blog.
I'm reviewing stuff from the aleks junk because it is the MOST COMPLICATED THING IN WORLD! I get it now though.
Simplifying a ratio of polynomials:
(w^2+6w-7) / (3w^2+33w+84)
To simply this expression you have to factor the numerator and denominator.
*first factor the top out:
w^2+7w-1w-7 = w(w+7)-1(w+7)
=(w-1)(w+7)
*now factor out the bottom:
3w^2+21w+12w+84 = 3w(w+7)+12(w+7)
=(3w+12)(w+7)
=3(w+4)(w+7)
=(w-1)(w+7) / 3(w+4)(w+7)
w+7 can cancel out so your left with
(w-1)/3(w+4)
and that's the final answer.
Simplifying a ratio of polynomials:
(w^2+6w-7) / (3w^2+33w+84)
To simply this expression you have to factor the numerator and denominator.
*first factor the top out:
w^2+7w-1w-7 = w(w+7)-1(w+7)
=(w-1)(w+7)
*now factor out the bottom:
3w^2+21w+12w+84 = 3w(w+7)+12(w+7)
=(3w+12)(w+7)
=3(w+4)(w+7)
=(w-1)(w+7) / 3(w+4)(w+7)
w+7 can cancel out so your left with
(w-1)/3(w+4)
and that's the final answer.
Nathan's blog
I really didn't understand finding the determinants of a 3 X 3 or a 4 X 4. That stuff just blew my mind. I am going to go over the section that is labeled as 3-D.
- absolute value of vector AB= √(x2-x1)^2+(y2-y1)^2+(z2+z1)^2
- midpoint= (x1+x2/2, y1+y2/2, z1+z2/2)
- Equation of a sphere - (x-x0)^2+(y-y0)^2+(z-z0)^2= r^2
- Vector equation - (x,y,z)=(x0,y0,z0)+ t(a,b,c)
- vector addition, magnitude, dot product, all follow same formula with extra variable.
Ex. 1 A sphere has points A(8,-2,3) and B(4,0,7) as endpoints of the diameter.
a. Find the center and radius
b. Find an equation of the sphere.
A.) Center= (8+4/2, 0+-2/2, 7+3/2)
c=6,-1,5
√(4-6)^2+(0-(-1)^2+(7-5)^2= 9
√2^2+1^2+2^2= √9= 3
B.) Equation of the sphere - (x-6)^2+(y-(-1)^2+(z-5)^2=9
So, if you know the formulas for this section it will be pretty easy. I will look over them a few more times before the test, so that I know that I have them memorized. That's all for this blog.
- absolute value of vector AB= √(x2-x1)^2+(y2-y1)^2+(z2+z1)^2
- midpoint= (x1+x2/2, y1+y2/2, z1+z2/2)
- Equation of a sphere - (x-x0)^2+(y-y0)^2+(z-z0)^2= r^2
- Vector equation - (x,y,z)=(x0,y0,z0)+ t(a,b,c)
- vector addition, magnitude, dot product, all follow same formula with extra variable.
Ex. 1 A sphere has points A(8,-2,3) and B(4,0,7) as endpoints of the diameter.
a. Find the center and radius
b. Find an equation of the sphere.
A.) Center= (8+4/2, 0+-2/2, 7+3/2)
c=6,-1,5
√(4-6)^2+(0-(-1)^2+(7-5)^2= 9
√2^2+1^2+2^2= √9= 3
B.) Equation of the sphere - (x-6)^2+(y-(-1)^2+(z-5)^2=9
So, if you know the formulas for this section it will be pretty easy. I will look over them a few more times before the test, so that I know that I have them memorized. That's all for this blog.
Friday, February 11, 2011
Taylor's 9th (I Think XD)blog Review!!
REVIEW TIME!!!
This is a review on my notes from lesson 8.1.
In this lesson we will be solving for THETA. In order for you to be able to solve for Theta you need to get the trig. function by itself and then you need to take an inverse of it. An inverse will have two answers, but there may be exceptions to this rule if the question asks you to just find one quadrant. Any way, you need to find where an angle is based on the trig function and then you need to tell if it is negative or positive.
These are the steps:
First, you need to take the inverse of the number so that you can find the First Quadrant angle.
Then you simply need to remember this in order to find the other answers:
Quadrant 2, you need to make the first quadrant angle a negative and add 180 degrees to it.
Quadrant 3, you just need to add 180 degrees to the first quadrant angle.
Quadrant 4, you need to make the first quadrant angle a negative and add 360 degrees to it.
Now let's solve some problems!!
(REMEMBER I DO NOT KNOW HOW TO PUT A GRAPH ON HERE SO PLEASE NOTE THIS. WHEN I SAY DRAW A QUARDNENT PLANE DO SO AND CHECK THE ONES THAT I TELL YOU TO OK!!)
Solve for Sin A= 2/5 for A.
Sin A = 2/5 First, write the equation down.
A= Sin^-1(2/5) Then put the equation in inverse form like I just did.
A= 23.578 degrees Then you find the inverse and make sure to round to the 3rd place after the decimal. Ok, now draw you quadrant plane. Now since our Sin is positive we need to put a check in the quadrants where Sin or Y would be positive. Which is the First and Second Quadrant. So since we have the First Quadrant already we just need to find the Second Quadrant.
-23.578 degrees + 180 degrees= 156.422 degrees Finally you just use the formula that I gave you to find the angle of the Second Quadrant and this is your answer.
So your answers should be 156.422 degrees and 23.578 degrees.
Let's try cosine next.
Solve for Cos H = 7/9 for H.
Cos H = 7/9 First, write the equation down.
H= Cos^-1(7/9) Then put the equation in inverse form like I just did.
H= 38.942 degrees Then you find the inverse and make sure to round to the 3rd place after the decimal. Ok, now draw you quadrant plane. Now since our Cos is positive we need to put a check in the quadrants where Cos or X would be positive. Which is the First and Fourth Quadrant. So since we have the First Quadrant already we just need to find the Fourth Quadrant.
-38.942 degrees + 360 degrees= 321.058 degrees Finally you just use the formula that I gave you to find the angle of the Fourth Quadrant and this is your answer.
So your answers should be 38.942 degrees and 311.81 degrees .
That is the review on my notes on Lesson 8.1.
This is a review on my notes from lesson 8.1.
In this lesson we will be solving for THETA. In order for you to be able to solve for Theta you need to get the trig. function by itself and then you need to take an inverse of it. An inverse will have two answers, but there may be exceptions to this rule if the question asks you to just find one quadrant. Any way, you need to find where an angle is based on the trig function and then you need to tell if it is negative or positive.
These are the steps:
First, you need to take the inverse of the number so that you can find the First Quadrant angle.
Then you simply need to remember this in order to find the other answers:
Quadrant 2, you need to make the first quadrant angle a negative and add 180 degrees to it.
Quadrant 3, you just need to add 180 degrees to the first quadrant angle.
Quadrant 4, you need to make the first quadrant angle a negative and add 360 degrees to it.
Now let's solve some problems!!
(REMEMBER I DO NOT KNOW HOW TO PUT A GRAPH ON HERE SO PLEASE NOTE THIS. WHEN I SAY DRAW A QUARDNENT PLANE DO SO AND CHECK THE ONES THAT I TELL YOU TO OK!!)
Solve for Sin A= 2/5 for A.
Sin A = 2/5 First, write the equation down.
A= Sin^-1(2/5) Then put the equation in inverse form like I just did.
A= 23.578 degrees Then you find the inverse and make sure to round to the 3rd place after the decimal. Ok, now draw you quadrant plane. Now since our Sin is positive we need to put a check in the quadrants where Sin or Y would be positive. Which is the First and Second Quadrant. So since we have the First Quadrant already we just need to find the Second Quadrant.
-23.578 degrees + 180 degrees= 156.422 degrees Finally you just use the formula that I gave you to find the angle of the Second Quadrant and this is your answer.
So your answers should be 156.422 degrees and 23.578 degrees.
Let's try cosine next.
Solve for Cos H = 7/9 for H.
Cos H = 7/9 First, write the equation down.
H= Cos^-1(7/9) Then put the equation in inverse form like I just did.
H= 38.942 degrees Then you find the inverse and make sure to round to the 3rd place after the decimal. Ok, now draw you quadrant plane. Now since our Cos is positive we need to put a check in the quadrants where Cos or X would be positive. Which is the First and Fourth Quadrant. So since we have the First Quadrant already we just need to find the Fourth Quadrant.
-38.942 degrees + 360 degrees= 321.058 degrees Finally you just use the formula that I gave you to find the angle of the Fourth Quadrant and this is your answer.
So your answers should be 38.942 degrees and 311.81 degrees .
That is the review on my notes on Lesson 8.1.
Charlie.
This week we did Aleks & we have been doing stuff with vectors.
vectors have a magnitude and a directions.
to add vectors you add head to the tail.
the triangle formed by the vectors give the resultant vector.
to add vectors algebraically you add components.
to find the magnitude > square root of x^2 + y^2
magnitude is done a|^->|
scalar multiplication puts a scalar times each component.
geometric representation = picture.
head ----tail
.________>
Example:
.__> A
.
|
|
|
|
\/ B
~~A+B =
.__>
|
|
|
|
\/
~~2A + 1/2B
.__.__>
|
|
\/
vectors have a magnitude and a directions.
to add vectors you add head to the tail.
the triangle formed by the vectors give the resultant vector.
to add vectors algebraically you add components.
to find the magnitude > square root of x^2 + y^2
magnitude is done a|^->|
scalar multiplication puts a scalar times each component.
geometric representation = picture.
head ----tail
.________>
Example:
.__> A
.
|
|
|
|
\/ B
~~A+B =
.__>
|
|
|
|
\/
~~2A + 1/2B
.__.__>
|
|
\/
Sunday, February 6, 2011
Feroz's Blog
Another long day doing Aleks stuff. Anyway, Go Packers.
I went to school 2 out of 5 days last week, so forgive me if I'm not sure what I'm talking about.
No one really went over 12-1, so I don't what that's all about. Something about lines. I don't know. So I guess that leaves Matrices. And I'm going to use 1x2's because anything bigger just looks ridiculous.
ex.
H = [2 3]
G = [4 5]
Find H + G
H + G = [6 8]
ex. 2 Find H x G
You can't really multiply them because they're dimensions don't work. The "inside" numbers have to match. ex. 2x3 3x4
*On a side note, Division requires you to multiply the matrix by the other's inverse. Again, that still requires the dimension's "inside" numbers to match.
I went to school 2 out of 5 days last week, so forgive me if I'm not sure what I'm talking about.
No one really went over 12-1, so I don't what that's all about. Something about lines. I don't know. So I guess that leaves Matrices. And I'm going to use 1x2's because anything bigger just looks ridiculous.
ex.
H = [2 3]
G = [4 5]
Find H + G
H + G = [6 8]
ex. 2 Find H x G
You can't really multiply them because they're dimensions don't work. The "inside" numbers have to match. ex. 2x3 3x4
*On a side note, Division requires you to multiply the matrix by the other's inverse. Again, that still requires the dimension's "inside" numbers to match.
Nathan's blog
Okay, so section 12-1 involved drawing little lines and connecting them. That's about it for that section. So, since that can't get me 150 words, I will just review matrices again. You can add, subtract, or multiply matrices but you can't divide them.
Adding Matrices:
[4 5] + [-9 2] = [-5 7]
[7 8] [5 -3] [12 5]
Subtracting Matrices:
[7 18] - [5 4] = [2 12]
[9 4] [-8 2] [17 2]
Multiplying Matrices:
[2 5] X [3 10] = [11 30]
[7 6] [1 2] [27 82]
That's only some of the things on matrices. You also have scalar multiplication, Cramer's rule, and finding the inverse and determinant of a matrix.
Cramer's rule is:
Dx/d
Dy/d
Scalar multiplication is when you solve without using Cramer's rule.
The determinant of a matrix is ad-bc.
The inverse of a matrix is 1/ the determinant of the matrix.
That's all I got on matrices. Next time I will probably review something different.
Adding Matrices:
[4 5] + [-9 2] = [-5 7]
[7 8] [5 -3] [12 5]
Subtracting Matrices:
[7 18] - [5 4] = [2 12]
[9 4] [-8 2] [17 2]
Multiplying Matrices:
[2 5] X [3 10] = [11 30]
[7 6] [1 2] [27 82]
That's only some of the things on matrices. You also have scalar multiplication, Cramer's rule, and finding the inverse and determinant of a matrix.
Cramer's rule is:
Dx/d
Dy/d
Scalar multiplication is when you solve without using Cramer's rule.
The determinant of a matrix is ad-bc.
The inverse of a matrix is 1/ the determinant of the matrix.
That's all I got on matrices. Next time I will probably review something different.
BLOOGGGGG
This week, we learned about matrices, which was reallllyyrealllyy realllyy easy whereas what i'm attemping to learn in aleks is driving me crazy. i mess up once and then fix it because it'll be an arithmetic error or i would have just typed it wrong on the last on before its "done" but then it says "oh, do it two more times". AHHHHHHHHHHHHHH. i'm gonna show you some of this maddness.
The thing i was stuck on was a number +/- a fraction OVER another number +/- another fraction! its so confusing!
but i don't have any problems to show you! i can make one up and show you the procedure, but i'm not gonna work it all the way through because its not gonna be pretty.
5-6x/2x^2
-------------
2+3/8x
say this is your problem, you have to find a commom denominator of the fractions inside the fraction. which would be 8x^2
then you multiply everything by this common denominator!
you would get:
45x^2-48x^3/2x^2
-------------------
16x^2+8x^2/8x
then you can reduce and get:
45x^2-24x
-----------
16x^2+x
then usually on aleks, it can be factored and reduced to a simple fraction, but i'm not gonna try mine, have fun once all ya'll get to that part of the aleks.:)
The thing i was stuck on was a number +/- a fraction OVER another number +/- another fraction! its so confusing!
but i don't have any problems to show you! i can make one up and show you the procedure, but i'm not gonna work it all the way through because its not gonna be pretty.
5-6x/2x^2
-------------
2+3/8x
say this is your problem, you have to find a commom denominator of the fractions inside the fraction. which would be 8x^2
then you multiply everything by this common denominator!
you would get:
45x^2-48x^3/2x^2
-------------------
16x^2+8x^2/8x
then you can reduce and get:
45x^2-24x
-----------
16x^2+x
then usually on aleks, it can be factored and reduced to a simple fraction, but i'm not gonna try mine, have fun once all ya'll get to that part of the aleks.:)
Kaitlyn's blogg
This week we did a test on matrices and we learned 12-1. We also worked on aleks a lot, which is good because its pretty easy and will help bring up everyone's grades in that class. I forgot how to do 12-1 and i forgot to bring my binder home, so i will go over matrices :)
Adding and subtracting matrices is pretty straightforward. you just add or subtract the number across from it in each matrix.
Ex:
[2 3] + [7 5] = [9 8]
[1 5] [1 0] [2 5]
When you multiply matrices, you must make sure the dimensions are correct. if you have a 3x2 matrix and a 2x1 matrix, you can multiply this because the two middle numbers (2 and 2) are the same.
Ex:
[2 3] x [7 5]= [14+ 3 10+0]= [17 10]
[1 5] [1 0] [7+5 5+0] [12 5]
Adding and subtracting matrices is pretty straightforward. you just add or subtract the number across from it in each matrix.
Ex:
[2 3] + [7 5] = [9 8]
[1 5] [1 0] [2 5]
When you multiply matrices, you must make sure the dimensions are correct. if you have a 3x2 matrix and a 2x1 matrix, you can multiply this because the two middle numbers (2 and 2) are the same.
Ex:
[2 3] x [7 5]= [14+ 3 10+0]= [17 10]
[1 5] [1 0] [7+5 5+0] [12 5]
lawrence's blog
there really isnt much to say about what we did this week. this week all we did was take a test on matrices and went over chapter 12-1. we also spent alot of time on aleks which is good because it will bring up alot of our grades. for those who didnt understand matrices it was pretty easy.
imma show yall if a matrix is able to be multiplied. for example:
3x2 2x2 you can multiply these because the last and first number are the same and you will have a 3x2
4x8 9x7 you cant multiply these because the first and last number arent the same
now imma add one just to be nice :)
2 5 7 + 4 5 6 = 6 10 13
3 9 1 6 7 8 9 16 9
that is all for now and if you need help with anything just let me know. i can help
imma show yall if a matrix is able to be multiplied. for example:
3x2 2x2 you can multiply these because the last and first number are the same and you will have a 3x2
4x8 9x7 you cant multiply these because the first and last number arent the same
now imma add one just to be nice :)
2 5 7 + 4 5 6 = 6 10 13
3 9 1 6 7 8 9 16 9
that is all for now and if you need help with anything just let me know. i can help
Friday, February 4, 2011
Taylor's Blog Review
REVIEW TIME!!!
These are the review notes on what we learned in section 7.5
You will need to know the following so that you can work out the equations.
Tan Theta=y/x
Cot Theta = x/y
Sec Theta = r/x
Csc Theta = r/y
Ok now that we know what Tan, Cot, Sec, and Csc are lets use them to solve the following equations.
Since Sin Theta = 2/8 0<> 90 degrees find the meaning of the other five trig functions.
(REMEMBER I DON’T KNOW HOW TO PUT A GRAPH ON THE COMPUTER SO I WILL JUST TYPE IT OUT.)
Ok, we know the since Sin is positive then the y needs to be positive and that can only happen on the second and first quadrant. Also since 2/8 is supposed to be less than 90 degrees we need to be in the first quadrant. Ok now we need to find x. You can draw a triangle and put in the info that you know to help you.
A^2+2^2=8^2
We will use the Pythagorean theorem to help us find x.
A^2+4=64
a^2=60
We will then use the exponents to get theses answers and then we will subtract 4 from 64 and get 60 then we will find the square root of a and 60 and the answer will be the square root of 60.
Now we just fill in the blanks.
Cos x/r= square root of 60/8
Tan y/x= (since we can’t have a square root at the bottom we will multiply it by the square root of 60 and reduce and get this) square root of 60/30
Cot x/y= square root of 60/2
Sec r/x=(since we can’t have a square root at the bottom we will multiply it by the square root of 60 and reduce and get this) 2 times the square root of 60/15
Csc r/y = 8/2 (reduced is) 4
Now lets try to find the reference angle of tan.
Find the reference angle for Tan 43 degrees.
First we find what quadrant its in. Which is the first because of the degrees which means tan is positive.
Tan 43 degrees
Since 65 degrees is already in-between 0 and 90 degrees we don’t need to subtract and since it’s not on the Trig chart that means that Tan 43 degrees is your answer.
That is 7.5 review Until next time!
These are the review notes on what we learned in section 7.5
You will need to know the following so that you can work out the equations.
Tan Theta=y/x
Cot Theta = x/y
Sec Theta = r/x
Csc Theta = r/y
Ok now that we know what Tan, Cot, Sec, and Csc are lets use them to solve the following equations.
Since Sin Theta = 2/8 0<> 90 degrees find the meaning of the other five trig functions.
(REMEMBER I DON’T KNOW HOW TO PUT A GRAPH ON THE COMPUTER SO I WILL JUST TYPE IT OUT.)
Ok, we know the since Sin is positive then the y needs to be positive and that can only happen on the second and first quadrant. Also since 2/8 is supposed to be less than 90 degrees we need to be in the first quadrant. Ok now we need to find x. You can draw a triangle and put in the info that you know to help you.
A^2+2^2=8^2
We will use the Pythagorean theorem to help us find x.
A^2+4=64
a^2=60
We will then use the exponents to get theses answers and then we will subtract 4 from 64 and get 60 then we will find the square root of a and 60 and the answer will be the square root of 60.
Now we just fill in the blanks.
Cos x/r= square root of 60/8
Tan y/x= (since we can’t have a square root at the bottom we will multiply it by the square root of 60 and reduce and get this) square root of 60/30
Cot x/y= square root of 60/2
Sec r/x=(since we can’t have a square root at the bottom we will multiply it by the square root of 60 and reduce and get this) 2 times the square root of 60/15
Csc r/y = 8/2 (reduced is) 4
Now lets try to find the reference angle of tan.
Find the reference angle for Tan 43 degrees.
First we find what quadrant its in. Which is the first because of the degrees which means tan is positive.
Tan 43 degrees
Since 65 degrees is already in-between 0 and 90 degrees we don’t need to subtract and since it’s not on the Trig chart that means that Tan 43 degrees is your answer.
That is 7.5 review Until next time!
Charlie's.
This week we reviewed & took a test on matrices [Chapter 14].
& like everyday we did aleks.com thingy.
(i still have like a wholeee bunch left to 'master')
& we just started chapter 12 yesterday.
chapter 12 is something weird.
i honestly already fergot how to do it.
the only thing i know is that you draw things,
& then if it says to add them, you connect the two arrows.
& on the two arrows thingy, there's a head and a tail.
so, since i totally fergot how to do it ALREADY.
i'll just do an example of matrices.
EXAMPLE:
[ 2 8 1 ] [ 8 2 7 ]
[ 6 5 7 ] + [ 4 6 9 ] =
[ 8 1 3 ] [ 4 6 2 ]
[ 10 10 8 ]
[ 10 11 16 ]
[ 12 7 5 ]
& like everyday we did aleks.com thingy.
(i still have like a wholeee bunch left to 'master')
& we just started chapter 12 yesterday.
chapter 12 is something weird.
i honestly already fergot how to do it.
the only thing i know is that you draw things,
& then if it says to add them, you connect the two arrows.
& on the two arrows thingy, there's a head and a tail.
so, since i totally fergot how to do it ALREADY.
i'll just do an example of matrices.
EXAMPLE:
[ 2 8 1 ] [ 8 2 7 ]
[ 6 5 7 ] + [ 4 6 9 ] =
[ 8 1 3 ] [ 4 6 2 ]
[ 10 10 8 ]
[ 10 11 16 ]
[ 12 7 5 ]
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