Free homework help here
- Thread starter MatthewGor123
- Start date
Thank ya.
Obviously (well, except 1^∞, ∞^0, 1/∞, but whatever)...but we're not talking about starting with infinity...Obviously, if you want to end up with infinity any moron could figure out that ∞ + 0, or 1 * ∞ is equal to infinity. What I *meant* to say was that without defining ∞ as a symbol, it is impossible to end up with infinity through the basic operations without stating a progression.
Thanks MatthewGor123, I wish I had a handy grandfather like yours. He's right, to solve the differential equation you need to "separate the variables". I'm still puzzled about how to do the entire questionBut me and my friends will work it out together when my holidays are over. Tell your granddad I say thanks.
I'm sorry I couldn't guide you closer to the answer, because I personally don't know calculus that well yet (I think I'm going to study it independently this year). I'm glad I could help at least a little
What class is this for? College level differential calculus?The Fat Controller
Guru
- Joined
- Aug 16, 2007
- Sythe Gold
- 1
- Messages
- 1,003
I'm doing further maths at my college and it was from the differential equations chapter.
Youtube is quite good if you want to study some maths independently, there are a bunch of videos up on pretty advanced concepts.
Youtube is quite good if you want to study some maths independently, there are a bunch of videos up on pretty advanced concepts.
Thanks for the tip! I would have never even considered YouTube as a very educational tool (like unless you want to learn how to do random things, like solve a Rubik's cube or some magic trick) but it does, in fact, have some pretty interesting stuff.
Now all I need is a few more questions :O
Now all I need is a few more questions :O
Alright, I guess I got unstickied. Oh well C'mon people, surely you must have SOME homework questions! I'm willing to answer them for any subject -> I have nothing better to do...
C'mon people, surely you must have SOME homework questions! I'm willing to answer them for any subject -> I have nothing better to do...
Darn, no new posts. I'm still willing to help For free!
For free!
I'm not going anywhere [assuming I don't get banned or something]. Ty for the post, anyways ^_^
[assuming I don't get banned or something]. Ty for the post, anyways ^_^
Okay, I need to find a reputable website that proves that blacks say the n-word more than non-blacks. Can you help?
1) http://www.streetgangs.com/magazine/053003niggas.php
2) http://www.thenewblackmagazine.com/view.aspx?index=37
I think these two sources could help you quite a bit, as they talk about both the culture of the N-word as well as cultural differences involved in its usage.
2) http://www.thenewblackmagazine.com/view.aspx?index=37
I think these two sources could help you quite a bit, as they talk about both the culture of the N-word as well as cultural differences involved in its usage.
madhacker14
Guru
Hey im a freshman, not real good at math just trying to keep up my 3.0 GPA, In my middle school we use to only use decimals, and they just automatically switched us to fractions, think you could PM, or post here a mini-guide on them, because i really need to get it quick, my teacher is senial, and has no idea how to teach.
Alright; sure. Are you asking how to convert decimals to fractions? Or like how to add/subtract/multiply them? What specifically do you want to know?
madhacker14
Guru
Alright;
ADDING FRACTIONS:
Let's say you want to add 3/8 and 9/14.
First thing you want to do is find the least common multiple of the denominators (the LCD). To find the LCD, unless you clearly recognize it, you write out the denominators in their prime factorization; 8 = 2 * 2 * 2, 14 = 2 * 7. Then, to find the LCD, you count the number of different prime factors each number has, and the LARGEST number of times each prime factor appears per number. If that made sense;
8 = 2^3
14 = 2^1 * 7^1.
Therefore, you need 3 2's and 1 7 to be factors of the LCD. Therefore, your LCD will be 2^3 * 7^1 = 56.
Now that you have your LCD, you have to manipulate your fractions so that the denominators of both are the same. So, for 3/8, you need to get 56 in the denominator (and we know it's a multiple of 8, as we just showed). Therefore, you can divide 56 by 8 to find how many times 8 goes into it. The answer is 7, so therefore, you rewrite 3/8 as (3*7)/(8*7) = 21/56.
Next, you do the same thing for 9/14. 56/14 = 4, so (9*4)/(14*4) = 36/56.
Next, you add 21/56 + 36/56. You add JUST the numerators, keeping the DENOMINATOR the same. (21 + 36) / 56 = 57/56. If you want, you can rewrite this as a mixed number; 1 and 1/56. Although it may seem quite confusing at first, it gets MUCH easier with practice.
~~~~~~~~~~~~~~~~~
SUBTRACTING Fractions
Let's say we wanna do 17/60 - 5/126.
Remember the first step? We have to find the LCD.
So;
60 = 2^2 * 3 * 5
126 = 2 * 3^2 * 7.
So, we add up the HIGHEST number of each prime factor in the denominators; we need two 2's, two 3's, a 5, and a 7 to make up the LCD. So, the LCD = 4 * 9 * 5 * 7 = 1260 (yikes).
But, we must solve the problem! 1260/60 = 21. 1260/126 = 10.
So; we have (17*21)/1260 - (10*5)/1260.
357/1260 - 50/1260 = 307/1260.
~~~~~~~~~~~~~~~~~~~~
MULTIPLYING Fractions
Luckily, here is where it gets easy
Let's say you want to multiply
3/4 * 8/11 * 33/21.
Generally, you want to cancel stuff (meaning divide by n/n, yielding 1) before you do the multiplication, but it doesn't really matter. I'd like to do that first. So, this expression can be rewritten as 8/4 * 33/11 * 3/21 (you still have 3 * 8 * 33 in the numerator and 4 * 11 * 21 in the denominator). So, you can rewrite this as 2 * 3 * 1/7 (since we know that 8/4 = 2 and 33/11 = 3). Therefore, the answer is 6/7.
However, a more straightforward way that leads you straight to the answer, then requires simplification, is just multiply the numerators and multiply the denominators. Literally. It's really simple. So, you have;
(3 * 8 * 33)/(4 * 11 * 21) = 792 / 924. Next, you want to simplify this. If you don't happen to know that both are divisible by 132, you can keep dividing by simple factors that go into both, i.e; 4 (198/231), then 3 (66/77), then 11 (6/7) and you'll end up with the same answer.
Quite simple really, once you get used to it.
~~~~~~~~~~~~~~~~
Let's say you want to divide 3/7 by 2/5.
So, you have (3/7) / (2/5).
First step is to flip the division sign to a multiplication sign, and find the reciprocal of the fraction on the RIGHT (the divisor). So, you'd have:
(3/7) * (5/2).
Then, you do your multiplication (which you just reviewed), and get 15/14, or 1 1/14.
Voila!
Basically, all it comes down to is practice, practice, and more practice. Hope this helped!
ADDING FRACTIONS:
Let's say you want to add 3/8 and 9/14.
First thing you want to do is find the least common multiple of the denominators (the LCD). To find the LCD, unless you clearly recognize it, you write out the denominators in their prime factorization; 8 = 2 * 2 * 2, 14 = 2 * 7. Then, to find the LCD, you count the number of different prime factors each number has, and the LARGEST number of times each prime factor appears per number. If that made sense;
8 = 2^3
14 = 2^1 * 7^1.
Therefore, you need 3 2's and 1 7 to be factors of the LCD. Therefore, your LCD will be 2^3 * 7^1 = 56.
Now that you have your LCD, you have to manipulate your fractions so that the denominators of both are the same. So, for 3/8, you need to get 56 in the denominator (and we know it's a multiple of 8, as we just showed). Therefore, you can divide 56 by 8 to find how many times 8 goes into it. The answer is 7, so therefore, you rewrite 3/8 as (3*7)/(8*7) = 21/56.
Next, you do the same thing for 9/14. 56/14 = 4, so (9*4)/(14*4) = 36/56.
Next, you add 21/56 + 36/56. You add JUST the numerators, keeping the DENOMINATOR the same. (21 + 36) / 56 = 57/56. If you want, you can rewrite this as a mixed number; 1 and 1/56. Although it may seem quite confusing at first, it gets MUCH easier with practice.
~~~~~~~~~~~~~~~~~
SUBTRACTING Fractions
Let's say we wanna do 17/60 - 5/126.
Remember the first step? We have to find the LCD.
So;
60 = 2^2 * 3 * 5
126 = 2 * 3^2 * 7.
So, we add up the HIGHEST number of each prime factor in the denominators; we need two 2's, two 3's, a 5, and a 7 to make up the LCD. So, the LCD = 4 * 9 * 5 * 7 = 1260 (yikes).
But, we must solve the problem! 1260/60 = 21. 1260/126 = 10.
So; we have (17*21)/1260 - (10*5)/1260.
357/1260 - 50/1260 = 307/1260.
~~~~~~~~~~~~~~~~~~~~
MULTIPLYING Fractions
Luckily, here is where it gets easy
Let's say you want to multiply
3/4 * 8/11 * 33/21.
Generally, you want to cancel stuff (meaning divide by n/n, yielding 1) before you do the multiplication, but it doesn't really matter. I'd like to do that first. So, this expression can be rewritten as 8/4 * 33/11 * 3/21 (you still have 3 * 8 * 33 in the numerator and 4 * 11 * 21 in the denominator). So, you can rewrite this as 2 * 3 * 1/7 (since we know that 8/4 = 2 and 33/11 = 3). Therefore, the answer is 6/7.
However, a more straightforward way that leads you straight to the answer, then requires simplification, is just multiply the numerators and multiply the denominators. Literally. It's really simple. So, you have;
(3 * 8 * 33)/(4 * 11 * 21) = 792 / 924. Next, you want to simplify this. If you don't happen to know that both are divisible by 132, you can keep dividing by simple factors that go into both, i.e; 4 (198/231), then 3 (66/77), then 11 (6/7) and you'll end up with the same answer.
Quite simple really, once you get used to it.
~~~~~~~~~~~~~~~~
Let's say you want to divide 3/7 by 2/5.
So, you have (3/7) / (2/5).
First step is to flip the division sign to a multiplication sign, and find the reciprocal of the fraction on the RIGHT (the divisor). So, you'd have:
(3/7) * (5/2).
Then, you do your multiplication (which you just reviewed), and get 15/14, or 1 1/14.
Voila!
Basically, all it comes down to is practice, practice, and more practice. Hope this helped!
Im stuck on this problem help
We are given unlimited number of identical, very thin dominoes. The length of a domino is 2 cm and it's mass is neglegable (10 ^ -4314 gram per domino).
Suppose we stack them(assume we are infinitely powerful), one on top of the other, so that the stack advances horizontally in one direction because each is offset a little. the top domino is offset 1cm, and as we move down the stack they are offset progressively less from each other so as to keep the center of gravity. the stack is kept stable only by gravity.
1) show that the stack can advance infinite distance. HINT: numerate the dominoes in a stack from the top. Note that a stack of two dominoes can advance 1cm since the center of the first domino should be above the edge of the second domino. Place the third domino underneath the first two dominoes so that the edge of the third domino is below the center of the stack of the two dominoes that has been built already.
We are given unlimited number of identical, very thin dominoes. The length of a domino is 2 cm and it's mass is neglegable (10 ^ -4314 gram per domino).
Suppose we stack them(assume we are infinitely powerful), one on top of the other, so that the stack advances horizontally in one direction because each is offset a little. the top domino is offset 1cm, and as we move down the stack they are offset progressively less from each other so as to keep the center of gravity. the stack is kept stable only by gravity.
1) show that the stack can advance infinite distance. HINT: numerate the dominoes in a stack from the top. Note that a stack of two dominoes can advance 1cm since the center of the first domino should be above the edge of the second domino. Place the third domino underneath the first two dominoes so that the edge of the third domino is below the center of the stack of the two dominoes that has been built already.
It seems pretty clear to me that it this is a calculus problem, and a difficult one at that, and it will involve a numeric progression - I'll try to think of what it may be, when I get a chance.
EDIT: I just did a quick web search regarding stacks with center of mass, and guess what I found! Somebody with the same exact problem as you!
http://answers.yahoo.com/question/index?qid=20061026223155AAXBOKnThat should explain it much better than I can, as the guy obviously knows his calculus better than I do.
If that wasn't sarcastic, then I really appreciate it
I look forward to trying my best to help 
Datsyuk4242
Forum Addict
- Joined
- Jul 18, 2006
- Sythe Gold
- 90
- Messages
- 528
Hmmm, just wondering if anyone could help me work out a little math problem. I calculated an answer that was ultimately different than what the textbook had, so it'd be great if I got a second opinion.
The members of a high school bastketball team are driving from Calgary to Vancouver, which is a distance of 675km. Each tire on their van has a radius of 32 cm. If the team members drive at a constant speed and cover the distance from Calgary to Vancouver in 6 H 45 min, what is the angular velocity, in radians/sec, of each tire during the drive?
Please and thank you
The members of a high school bastketball team are driving from Calgary to Vancouver, which is a distance of 675km. Each tire on their van has a radius of 32 cm. If the team members drive at a constant speed and cover the distance from Calgary to Vancouver in 6 H 45 min, what is the angular velocity, in radians/sec, of each tire during the drive?
Please and thank you
- Joined
- Jul 28, 2008
- Sythe Gold
- 5
- Messages
- 869
If that wasn't sarcastic, then I really appreciate it
Nah man, wasn't sarcastic.
Hmmm, just wondering if anyone could help me work out a little math problem. I calculated an answer that was ultimately different than what the textbook had, so it'd be great if I got a second opinion.
The members of a high school bastketball team are driving from Calgary to Vancouver, which is a distance of 675km. Each tire on their van has a radius of 32 cm. If the team members drive at a constant speed and cover the distance from Calgary to Vancouver in 6 H 45 min, what is the angular velocity, in radians/sec, of each tire during the drive?
Please and thank you
I think you would take 32cm as your radian measure and then divide that by 6 h and 45mins converted to secs. This should give you rads/sec
Similar threads
