Saturday, August 22, 2009
newer study group
yeah i say one at my house 2 days before the test since brandi thinks that would be a better idea.
REFLECTION #1
I'm glad to say that Chapter 1 this week wasn't as bad as I thought it would be. I thought it would be hard to remember what I learned from last year, but as we reviewed the different things, they became easier than before. The more I did the same types of problems, the more I understood the concept and the less time it took me to work the problems.
For the most part, I pretty much understand everything that we learned in Chapter 1, but the easiest thing for me to explain is simple factoring equations.
For example: x^2 + 3x + 2.
This problem is really easy. The first thing you have to do is ask yourself, "What are the factors of 2 that can add to give me 3?" You would use the numbers, 2 and 1 and they would both be positive. (because the back sign and front sign in the equation are +) Then you would get (x + 2) and (x + 1). Then you would simply set (x + 2)=0 and get -2, and set (x + 1)=0 and get -1. Lastly, you would put those numbers in point form => so your final answers would be (-2,0) and (-1,0).
Another example would be this: x^2 - 3x = 0
Alright, this equation looks hard because there is no constant, but it's really not. The first thing you can do is factor out an x. And you would get x(x-3)=0. Then you just set x=0 and set x-3=0 and you would get the numbers 0 and 3. And after you put them in point form your answer should look like this => (0,0) (3,0).
Okay so the only thing from Chapter 1 that I really don't understand is problem #18 on our test. => "Find the reciprocal of i and simplify." This problem confused me a lot because i thought the reciprocal of i would be the square root of 1 and that would simplify to equal 1. But I was incorrect. Since i is the same thing as the square root of -1 this problem stumped me. **So if anyone knows how to work this problem and would like to explain it to me that would be great :)
Overall, I thought Chapter 1 was pretty easy basically because I learned this last year.
For the most part, I pretty much understand everything that we learned in Chapter 1, but the easiest thing for me to explain is simple factoring equations.
For example: x^2 + 3x + 2.
This problem is really easy. The first thing you have to do is ask yourself, "What are the factors of 2 that can add to give me 3?" You would use the numbers, 2 and 1 and they would both be positive. (because the back sign and front sign in the equation are +) Then you would get (x + 2) and (x + 1). Then you would simply set (x + 2)=0 and get -2, and set (x + 1)=0 and get -1. Lastly, you would put those numbers in point form => so your final answers would be (-2,0) and (-1,0).
Another example would be this: x^2 - 3x = 0
Alright, this equation looks hard because there is no constant, but it's really not. The first thing you can do is factor out an x. And you would get x(x-3)=0. Then you just set x=0 and set x-3=0 and you would get the numbers 0 and 3. And after you put them in point form your answer should look like this => (0,0) (3,0).
Okay so the only thing from Chapter 1 that I really don't understand is problem #18 on our test. => "Find the reciprocal of i and simplify." This problem confused me a lot because i thought the reciprocal of i would be the square root of 1 and that would simplify to equal 1. But I was incorrect. Since i is the same thing as the square root of -1 this problem stumped me. **So if anyone knows how to work this problem and would like to explain it to me that would be great :)
Overall, I thought Chapter 1 was pretty easy basically because I learned this last year.
Reflection 1
Alright, I have a question it probably wont make much sense cause I don't know how to explain it on the computer but i'm so lost so someone please try to help me. I need help with problems 17 and 18 on the homework we got Friday. Section 2.1 page 56.
Number 17 is f(x)= x^3-9x
a. f(-square root of 2/3) b. f(i square root of 3) c. f(x/3) d. f(x-3)
I thought when you had problems like those you were suppose to do the thing where you press y= on your calculator and then press the graph button. I tried doing it and it gave me huge decimal numbers so I figured I was wrong. So then I tried plugging the number it gives you in a, b, c, and d into f(x) but I didn't know what to do after that.
One thing I understood was is in Chapter one, Conjugates. There's not much to say about it because it's pretty simple. All you do is give the opposite sign of the denominator.
Example: 5-2i/4+3i
The conjugate would be 4-3i
If it would ask you to put it into a+bi form all you do is multiply the numerator and denominator by the conjugate. FOIL the top and the bottom, then put your answer into a+bi form.
Example: 5-2i/4+3i x 4-3i/4-3i=14-23i/25
=14/25 + -23/25i
Number 17 is f(x)= x^3-9x
a. f(-square root of 2/3) b. f(i square root of 3) c. f(x/3) d. f(x-3)
I thought when you had problems like those you were suppose to do the thing where you press y= on your calculator and then press the graph button. I tried doing it and it gave me huge decimal numbers so I figured I was wrong. So then I tried plugging the number it gives you in a, b, c, and d into f(x) but I didn't know what to do after that.
One thing I understood was is in Chapter one, Conjugates. There's not much to say about it because it's pretty simple. All you do is give the opposite sign of the denominator.
Example: 5-2i/4+3i
The conjugate would be 4-3i
If it would ask you to put it into a+bi form all you do is multiply the numerator and denominator by the conjugate. FOIL the top and the bottom, then put your answer into a+bi form.
Example: 5-2i/4+3i x 4-3i/4-3i=14-23i/25
=14/25 + -23/25i
2nd Try. Reflection 1
Okay to start off i would like to say wow! This week we moved really fast in class with reviewing, and the formulas were difficult to learn, having to learn them all at once. B-Rob was right with doing all of our homework because by the time for our Chapter 1 test I understood how to work most of the problems. I do know to definately ask if i have a question because it will help when you have right answer to study instead of the wrong one, and also actually understanding how the problem is worked. With all the homework and formulas, the first week was sorta overwelming, but hopefully things will get easier as I get used to the class. One of the lessons i found easiest to work and had easier formulas to memorize was factoring, completing the square, and quadratic formula. The most difficult lesson that i had more trouble learning was finding the perpendicular bisector of a line. So, since i grasped the idea of completing the square the easiest, I'll give an example of one of the problems from our Chapter 1 test:
13.)Solve for x by completing the square x^2 - 4x = 9
First you must dived b by 2, then square the answer of b divided by 2 then plug it in to the eqn.
-4/2 = (-2)^2 = 4 x^2-4x+4=9+4 Second you take the -2 from parenthesis and square it minus x, also must add 9+4 on the other side of equal sign. (x-2)^2=13 Third, you take the square root.
square root of (x-2)^2= square root of 13. Next you add 2 to each side giving you: x=2+ 0r - square root of 13. and Finally put the answer in point form: (2+squareroot of 13, 0) (2-squareroot of 13, 0).
Overall, I did think this week was some what difficult. But hopefully i will get the hang of things and keep improving in class.
13.)Solve for x by completing the square x^2 - 4x = 9
First you must dived b by 2, then square the answer of b divided by 2 then plug it in to the eqn.
-4/2 = (-2)^2 = 4 x^2-4x+4=9+4 Second you take the -2 from parenthesis and square it minus x, also must add 9+4 on the other side of equal sign. (x-2)^2=13 Third, you take the square root.
square root of (x-2)^2= square root of 13. Next you add 2 to each side giving you: x=2+ 0r - square root of 13. and Finally put the answer in point form: (2+squareroot of 13, 0) (2-squareroot of 13, 0).
Overall, I did think this week was some what difficult. But hopefully i will get the hang of things and keep improving in class.
Sample Reflection
Here is a sample reflection. Notice how he mentions something he didn't understand and explains something he did. He doesn't ask a question though. That is the only thing missing from this. Your reflection must be a minimum of 250 words.
Sample:
So I want to start this post off with saying that this week was not what I expected it to be. I expected the first week to be really hard, especially after seeing ALL of the formulas for the derivatives. However, it turned out not to be as intimidating as I thought. I think I really grasped hold on the concepts behind taking a derivative rather well. I guess it helps that through Mu Alpha Theta I had already learned the basic derivative formula...the whole limit as h approaches 0 thing, as well as taking a derivative of a polynomial the short way. I guess I was just more scared about Calculus because I know that I really need to focus on doing everything right and understanding it completely in order to do well on the AP exam...
Firstly, I think the only thing I didn't grasp at first was the weird wording on the word problems. There seems to be so many different words for the same exact thing, and it just gets really confusing. I think the only thing I had problems with on the packet was the average speed thing...and I think I figured out why. In our notes, we used an example problem with "the first two seconds". I did not realize at first that it doesn't have to be the "first nseconds" it can be any interval, such as from 2 to 4. It was also kind of confusing there because we used the long way to find the derivative, and I got lost in my notes...I guess I should learn to take better notes in math.
One thing that I really like the feel of that I think I could help you guys with is when taking care of things like:
1
---
x(2/3)
I think everyone gets the concept of bringing it back up but I'm more than sure that a lot of you guys are forgetting the derivative rule when doing this. You bring the exponent to the front and then SUBTRACT one. You have to remember the subtract. I don't know how many times you guys are telling me that -(2/3) - 1 = (1/3). It becomes -(5/3), and therefore the answer would be
-2
---
3x(5/3)
I hope that clears up some problems for some of you guys...
Anyway, as far as our first week, I think we are doing strong as a class, and we can move forward next week even stronger...We are doing really well so keep up the good work guys!
Be ready for the quiz on Tuesday :-p
Firstly, I think the only thing I didn't grasp at first was the weird wording on the word problems. There seems to be so many different words for the same exact thing, and it just gets really confusing. I think the only thing I had problems with on the packet was the average speed thing...and I think I figured out why. In our notes, we used an example problem with "the first two seconds". I did not realize at first that it doesn't have to be the "first nseconds" it can be any interval, such as from 2 to 4. It was also kind of confusing there because we used the long way to find the derivative, and I got lost in my notes...I guess I should learn to take better notes in math.
One thing that I really like the feel of that I think I could help you guys with is when taking care of things like:
1
---
x(2/3)
I think everyone gets the concept of bringing it back up but I'm more than sure that a lot of you guys are forgetting the derivative rule when doing this. You bring the exponent to the front and then SUBTRACT one. You have to remember the subtract. I don't know how many times you guys are telling me that -(2/3) - 1 = (1/3). It becomes -(5/3), and therefore the answer would be
-2
---
3x(5/3)
I hope that clears up some problems for some of you guys...
Anyway, as far as our first week, I think we are doing strong as a class, and we can move forward next week even stronger...We are doing really well so keep up the good work guys!
Be ready for the quiz on Tuesday :-p
Friday, August 21, 2009
reflections 1
so anybody wanna help me with 19. on the chapter 1 test? i have no clue how to find the equation of a bisector of the line, it confuses me with points and how it has to be through a certain point. >_< help
ok, so....
what are we supposed to post?
is it one question we dont get?
and then 2 comments on other ppls posts?
someone help
Reflections 1
Can someone help me work problems 10 & 19 on the adv. math chapter 1 test.
10.) What is the equation of the horizontal line through ( -2, 3).
19.) Find the equation of the perpendicular bisector of the line 4x-2y=4 through the point ( 1, 4).
10.) What is the equation of the horizontal line through ( -2, 3).
19.) Find the equation of the perpendicular bisector of the line 4x-2y=4 through the point ( 1, 4).
Thursday, August 20, 2009
Wednesday, August 19, 2009
Tuesday, August 18, 2009
Can somebody help me out?
Page 41. As seen in numbers 1.c and 17, I keep coming out with imaginary roots. I can't figure out how to graph them? For #17, i tried completing the square. So, before i square rooted both sides, i have (x+1)^2 = -16. then i square rooted each side, and got x + 1 = 4i, therefore x = -1 ± 4i. Does anybody know how to graph the imaginary roots on the x-axis, or is it possible? I had the same problem with #1.c and #22. I don't know, maybe i did it wrong, can somebody double check? Thanks.
Monday, August 17, 2009
now hopefully someone can enlighten me....
when i was trying to help Matt i realized that i did number 12 wrong at first, so i went back and ran through the steps once more and i was able to help him after. Goes to show what happens when you try and rush! haha.
anyways, i'm stuck on Factoring, back pg. numbers 20, 25,26,28.
On 25 and 28 i got far enough to realize that they are the same problem once you remove the GCF from 26 but since i was already stuck on 25, so that wasn't too beneficial.
on 26, how can you factor a quadratic with the linear coefficient and the back number the same?
i.e.,
x^2+4x+4=0
because the leading coeff is 1, i can't factor out a GCF, so where do i go from here?
if i were to factor my answer would be 3 or 5.
(x+1)(x+4)<-- or vice versa....
On number 20 i'm not sure what to do being that it's not a quadratic, i tried factoring out a square but it's the square root of one, is that possible?!
also, on 25&28 both problems have 1 as the linear coefficient for x and how is that factorable?
if anyone can give me at least the first step, i'm pretty sure i'll be able to go from there.
please and thanks!
anyways, i'm stuck on Factoring, back pg. numbers 20, 25,26,28.
On 25 and 28 i got far enough to realize that they are the same problem once you remove the GCF from 26 but since i was already stuck on 25, so that wasn't too beneficial.
on 26, how can you factor a quadratic with the linear coefficient and the back number the same?
i.e.,
x^2+4x+4=0
because the leading coeff is 1, i can't factor out a GCF, so where do i go from here?
if i were to factor my answer would be 3 or 5.
(x+1)(x+4)<-- or vice versa....
On number 20 i'm not sure what to do being that it's not a quadratic, i tried factoring out a square but it's the square root of one, is that possible?!
also, on 25&28 both problems have 1 as the linear coefficient for x and how is that factorable?
if anyone can give me at least the first step, i'm pretty sure i'll be able to go from there.
please and thanks!
Factoring
I know its easy but some of the factoring problems are giving me trouble on the worksheet. Let's see #12: -4x2-14x+8=0 is one of the ones killing me. I'm sure if someone can help me with this one it'll help me with the other ones i'm not understanding. Thanks.
Sunday, August 16, 2009
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