If you have a math problem you need help with, I'll give help.
Rules:
- I won't help you with more than one of the same type of math problem; this is to avoid doing your homework for you.
- I will discuss a problem I've explained, but will not reattempt to demonstrate the same method on another problem.
- I expect you to be able to do the math I explain, once I do so.
- Don't give me trivial math problems designed to waste my time. If you do so, I shall ask a moderator to delete the post. With luck and cooperation, this shouldn't happen too much
- I can't help with anything as complicated or more so than integral calculus.
Guidelines:
- If you have a question, post it in a separate post from the one where you discussed the previous problem, if you did. It's not a double post, since the subject matters are completely different, and the post is at a later time.
- If you'd like to discuss a problem, post only once after each problem. If you have something more to say, modify it in. After a few days, if the conversation is over or nearly over, I'll put the conversation in a spoiler, where one can go back and read it. Once I've done that, I'll ask people who participated in the conversation to delete their posts. That way, I maintain a clean question-answer format.
I just wanted to say that I think that, sometimes, the doubts raised in the discussion may be interesting for others to understand the result...
(If you want, I can delete this post.)
By the way, can we ask you to give us things like an exact formula for the perimeter of an ellipse? :P
This is one I've seen floating around facebook recently, and I thought it had the potential of being open to interpretation. It depends on how you interpret the older division symbol, and what priority you consider the multiplication by parentheses to have. I know what I think is the correct answer. I just want to see your response.
6÷2(1+2)=?
Quote from: MikeW781 on April 30, 2011, 09:43:53 PM
This is one I've seen floating around facebook recently, and I thought it had the potential of being open to interpretation. It depends on how you interpret the older division symbol, and what priority you consider the multiplication by parentheses to have. I know what I think is the correct answer. I just want to see your response.
6÷2(1+2)=?
It equals 9, because, in the order of operations, parenthesis come first, so, 2+1=3, now it's 6/2x3, so, now you go left to right, 6/2=3, 3x3=9. Sorry for stealing the show, DD.
I came upon the same answer.
Quote from: Duskling on April 30, 2011, 09:52:06 PM
Quote from: MikeW781 on April 30, 2011, 09:43:53 PM
This is one I've seen floating around facebook recently, and I thought it had the potential of being open to interpretation. It depends on how you interpret the older division symbol, and what priority you consider the multiplication by parentheses to have. I know what I think is the correct answer. I just want to see your response.
6÷2(1+2)=?
It equals 9, because, in the order of operations, parenthesis come first, so, 2+1=3, now it's 6/2x3, so, now you go left to right, 6/2=3, 3x3=9. Sorry for stealing the show, DD.
The thing is, that some people put forth the argument that multiplication by parentheses took priority, so one would multiply the 2 by the (3) before dividing, resulting in a final answer of six.
Also, others interpreted the division symbol as applying to both the 2 and the (1+2), so it would look like:
__
6__
2(1+2)
While I agree that the answer is 9 based of what I've been taught, I was curious to see the point of view of somebody else who had been taught either of the above methods for solving this.
Quote from: MikeW781 on May 01, 2011, 01:59:40 PM
Quote from: Duskling on April 30, 2011, 09:52:06 PM
Quote from: MikeW781 on April 30, 2011, 09:43:53 PM
This is one I've seen floating around facebook recently, and I thought it had the potential of being open to interpretation. It depends on how you interpret the older division symbol, and what priority you consider the multiplication by parentheses to have. I know what I think is the correct answer. I just want to see your response.
6÷2(1+2)=?
It equals 9, because, in the order of operations, parenthesis come first, so, 2+1=3, now it's 6/2x3, so, now you go left to right, 6/2=3, 3x3=9. Sorry for stealing the show, DD.
The thing is, that some people put forth the argument that multiplication by parentheses took priority, so one would multiply the 2 by the (3) before dividing, resulting in a final answer of six.
Also, others interpreted the division symbol as applying to both the 2 and the (1+2), so it would look like:
__6__
2(1+2)
While I agree that the answer is 9 based of what I've been taught, I was curious to see the point of view of somebody else who had been taught either of the above methods for solving this.
This is what I thought at first, until one of my friends explained it to me.
I would disagree.
| 6/2(1 + 2) = x | Given |
| a/b(1+2) = x | Let a = 6
Let b = 2
|
| a/b(3) = x | Addition |
| a/bc = x | Let c = 3 |
| (a) / (bc) = x | Terms cannot be separated by order of operations. a/bc is not equal to ab-1c, but rather to ab-1c-1 |
| (6) / (2*3) = x | Substitution of symbols |
| 6 / 6 = x | Multiplication and removal of parentheses |
| x = 1 | Division |
I think that where you got lost was where you made the mistake of equating bc and b*c. b*c can be separated if the group of symbols is divided. Then, only the b is inverted. However, bc is a separate term, and cannot be separated using the hierarchy of left-to-right.
Uh, I don't think I'm gonna be here much if that's the form answers are going to be in.
I don't really think that you covered all the details of the subject, Demon Duck.
The main problem here is that writing ab/cd isn't following the standard for math notation (at least, as far as I know).
So, when we write ab/cd do we mean:
(a.b)/(c.d), where "." means the multiplication period, i.e.,
Or does ab/cd means:
a.(b/c).d, i.e.,
=a.b.d/c
If we're programming, when we write a*b/c*d (with "*" meaning multiplication), this is equal to a*b*d/c. If I want to have the first case, I'll have to write a*b/(c*d) or a*b/c/d.
By the way, usually the computation of a*b/(c*d) is slightly faster and uses a bit more memory than a*b/c/d.
Quote from: Demon Duck on May 01, 2011, 09:32:38 PM
I would disagree.
This is incorrect. a/bc = a/b*c= a/b THEN *c
Basically, you added parentheses when you were not supposed to.
I can't stop laughing, this would be shown in a textbook to be 9 end of story. I find this interesting but more than anything funny!
Quote from: The Holy namelesskitty on May 02, 2011, 06:02:14 PM
I can't stop laughing, this would be shown in a textbook to be 9 end of story. I find this interesting but more than anything funny!
Well, yeah, it is sad. On the facebook poll millions of people picked 1
Quote from: MikeW781 on May 02, 2011, 07:12:56 PM
Quote from: The Holy namelesskitty on May 02, 2011, 06:02:14 PM
I can't stop laughing, this would be shown in a textbook to be 9 end of story. I find this interesting but more than anything funny!
Well, yeah, it is sad. On the facebook poll millions of people picked 1
Not sad, I'd just say they have a different view on it, is all, I think picking 1, while incorrect, is perfectly rational.
Yeah, I'm not sure where the one came from, but in any case it's interesting to see different viewpoints.
I've never seen such a textbook, Cat.
How does one pronounce the term "a/bc"?
How does one pronounce the term "a/b*c"?
In one, the term "bc" is pronounced as a single term, multiplied through juxtaposition, which, I believe, takes precedence over other modes of multiplication and division. The values in a term multiplied by juxtaposition cannot be split apart by a division symbol with no other grouping marks.
Everyone:
6/2(2 + 1)
is not equal to:
6
2(2+1)
The way that the question is written (6/2(2 + 1)), it = 9. This is because when you have Division and Multiplication (in the same order) you go from left to right.
However, if the question were written in the way below, it would = 1 because their are inferred brackets around the top and bottom.
[spoiler=Discussion][spoiler=Demon Duck answer]
I would disagree.
| 6/2(1 + 2) = x | Given |
| a/b(1+2) = x | Let a = 6
Let b = 2
|
| a/b(3) = x | Addition |
| a/bc = x | Let c = 3 |
| (a) / (bc) = x | Terms cannot be separated by order of operations. a/bc is not equal to ab-1c, but rather to ab-1c-1 |
| (6) / (2*3) = x | Substitution of symbols |
| 6 / 6 = x | Multiplication and removal of parentheses |
| x = 1 | Division |
I think that where you got lost was where you made the mistake of equating bc and b*c. b*c can be separated if the group of symbols is divided. Then, only the b is inverted. However, bc is a separate term, and cannot be separated using the hierarchy of left-to-right.
[/spoiler]
[spoiler=Duskling]
It equals 9, because, in the order of operations, parenthesis come first, so, 2+1=3, now it's 6/2x3, so, now you go left to right, 6/2=3, 3x3=9. Sorry for stealing the show, DD.[/spoiler]
[spoiler=Idozen Cair]I came upon the same answer.[/spoiler]
[spoiler=MikeW781]The thing is, that some people put forth the argument that multiplication by parentheses took priority, so one would multiply the 2 by the (3) before dividing, resulting in a final answer of six.
Also, others interpreted the division symbol as applying to both the 2 and the (1+2), so it would look like:
__6__
2(1+2)
While I agree that the answer is 9 based of what I've been taught, I was curious to see the point of view of somebody else who had been taught either of the above methods for solving this.[/spoiler]
[spoiler=Duskling]This is what I thought at first, until one of my friends explained it to me.[/spoiler]
[spoiler=Bugfartboy]Uh, I don't think I'm gonna be here much if that's the form answers are going to be in.[/spoiler]
[spoiler=Ertxiem]I don't really think that you covered all the details of the subject, Demon Duck.
The main problem here is that writing ab/cd isn't following the standard for math notation (at least, as far as I know).
So, when we write ab/cd do we mean:
(a.b)/(c.d), where "." means the multiplication period, i.e.,
Or does ab/cd means:
a.(b/c).d, i.e.,
=a.b.d/c
If we're programming, when we write a*b/c*d (with "*" meaning multiplication), this is equal to a*b*d/c. If I want to have the first case, I'll have to write a*b/(c*d) or a*b/c/d.
By the way, usually the computation of a*b/(c*d) is slightly faster and uses a bit more memory than a*b/c/d.[/spoiler]
[spoiler=MikeW781]
Quote from: Demon Duck on May 01, 2011, 09:32:38 PM
I would disagree.
This is incorrect. a/bc = a/b*c= a/b THEN *c
Basically, you added parentheses when you were not supposed to.[/spoiler]
[spoiler=Holy Cat]I can't stop laughing; this would be shown in a textbook to be 9 -- end of story. I find this interesting but, more than anything, funny!
[/spoiler]
[spoiler=MikeW871]Well, yeah, it is sad. On the facebook poll millions of people picked 1.
[/spoiler]
[spoiler=Duskling]
Not sad, I'd just say they have a different view on it, is all, I think picking 1, while incorrect, is perfectly rational.[/spoiler]
[spoiler=Holy Cat]Yeah, I'm not sure where the one came from, but in any case it's interesting to see different viewpoints.[/spoiler]
[spoiler=Demon Duck]I've never seen such a textbook, Cat.
How does one pronounce the term "a/bc"?
How does one pronounce the term "a/b*c"?
In one, the term "bc" is pronounced as a single term, multiplied through juxtaposition, which, I believe, takes precedence over other modes of multiplication and division. The values in a term multiplied by juxtaposition cannot be split apart by a division symbol with no other grouping marks.
[/spoiler]
[spoiler=Zack]Everyone: I think
6/2(2 + 1) is not equal to:
6
2(2+1)The way that the question is written (
6/2(2 + 1)), it = 9. This is because when you have Division and Multiplication (in the same order) you go from left to right.
However, if the question were written in the way below, it would = 1 because their are inferred brackets around the top and bottom.
[/spoiler]
[/spoiler]
Okay... does anyone have a math problem they
don't already know the answer to?
Yes.
2x^(2)*4.8-3=x
What is x?
(I put the exponent in paranthesis to avoid confusing people into thinking it's x^2*4.8 instead of x^2 multiplied by4.8)
2x^(2)*4.8-3=x
4x^2*4.8-3=x
19.2x^2-3=x
19.2x^2-x-3=0
x=1+sqrt(1-4*19.2*-3)
38.4
x=approx 0.422183268
OR
x=1-sqrt(1-4*19.2*-3)
38.4
x=approx -0.370099935
Quote from: Bugfartboy on May 08, 2011, 08:43:10 AM
Yes.
2x^(2)*4.8-3=x
What is x?
(I put the exponent in paranthesis to avoid confusing people into thinking it's x^2*4.8 instead of x^2 multiplied by 4.8 )
Do you mean
a) 2x
2 * 4.8 - 3 = x (http://www.wolframalpha.com/input/?i=2x%5E%282%29*4.8-3%3Dx)
or
b) (2x)
2 * 4.8 - 3 = x (http://www.wolframalpha.com/input/?i=%282x%29%5E2*4.8-3%3Dx)
I think you meant a), but Shadoroq solved b).
Solving a) gives
9.6 x
2 - x - 3 = 0 ⇔
x = (1 ± √(1 + 4*9.6*3)) / (2*9.6) ⇔
x = (1 ± √(1 + 4*9.6*3)) / (2*9.6) ⇔
x ≈ -0.51 ∨ x ≈ -0.61
By the way, the [ sup ] 2 [ /sup ] is another way of making powers.
And to avoid smileys I added a space between the "8" and the ")".
And sorry, Demon Duck, for taking you the spotlight and linking to WolframAlpha. :)
4.8(2x2) - 3 = x
Now, the 2 coefficient of the x2 isn't squared, so the equation can be rephrased as
(4.8 × 2 × x2) - 3 = x
Subtracting x from both sides, and multiplying the two coefficients of x2, one gets
9.6x2 - x - 3 = 0
The quadratic formula tells us the rest. If a = 9.6, b = -1, and c = -3, then
-b ± (b2 - 4ac)0.5
x = 2a . Substituting in the values gives us
-(-1) ± ((-1)2 - 4(9.6)(-3))0.5
x = 2(9.6) . Through a couple of steps of simplification,
1 ± (1 - (-115.2))0.5
x = 19.2 , and then
1 ± (116.2)0.5
x = 19.2 . Now, if you want the fractional and root answer,
1 sqrt(116.2)
x = ±
19.2 19.2 . This can also be expressed as
5 sqrt(116.2) 5 2905
x = ± = ± sqrt [ ]
96 sqrt(368.64) 96 9216 . If you want the decimal form, it's approximately
x ≈ 0.05208333... ± 0.5614380 ≈ 0.6135214 OR -0.509355
Somehow, my answer differs from Ert's, which means I'm probably wrong. If not, I was glad to help.
EDIT: Thanks, Ert.
Wanna know something? I really just made that up on the spot so you don't get out of practice. But now I know the answer to a math question that I hopefully will never get. :D
Demon Duck: Can we ask you harder stuff, like the eigenvalues of a tridiagonal matrix, or the probability of having three of a kind when randomly picking up 5 cards, or an iterative formula that converges exponentially to
π, or can you give us a mathematical process to draw plants or snowflakes, or an infinite line drawn in a limited space, or... sorry, I seem to have drifted away on my own... :) And on an only apparently unrelated subject: did you know that I made the code to colour the background of my avatar, can you recognize it (see the attachment)?
Psst, another thing, Demon Duck.
Well, the chances of three of a kind:
There are different possibilities to picking up certain cards. Each card value is designated as a letter. One can pick up five different ones:
ABCDE -- this has a probability of 0.5070828, the product of 48/51, 44/50, 40/49, and 36/48. One card is removed from the denominator each time since one less card remains, but four are removed from the numerator each time.
AABCD, ABACD, ABCAD, ABCDA, ABBCD, ABCBD, ABCDB, ABCCD, ABCDC, or ABCDD -- all of these have probabilities of 0.0422569. This is the product of 48, 44, 40, and 3 as the numerator (in some order), as different numerical values diminish, and you need one of the three cards left in the value to get the pair. 51, 50, 49, and 48 are the factors of the numerator.
AABBC, AABCB, AABCC, ABABC, ABACB, ABACC, ABBAC, ABCAB, ABCAC, ABBCA, ABCBA, ABCCA, ABBCC, ABCBC, or ABCCB -- each of these 15 has a probability of 0.003169268. Summed, this is the probability of being dealt two-pair.
Any hand where the cards are not all different, single-paired, or double-paired, there is at least a three of a kind in the hand. Thus, the (poorly) calculated probability of getting three-of-a-kind from any five cards should be 1 - .5070828 - (10 * .0422569) - (15 * .003169268); this equals . The probability of getting three of a kind from any five cards drawn should be about 25 in 1096, or lightly better than one in 44.
Yes, you can post the sort of stuff like that which was about pi, and the cards (simple probability problem), and the infinite line in limited space is impossible; it has to be a curve as far as I know, or the space has to be curved. Just don't ask me much about linear algebra and matrices.
Well, actually, I asked for the probability of having 3 of a kind. This is 0.02113... or about 3 in 142. Yeah... I knew the answer... sorry...
A poker (4 of a kind) is not included, so it should also be considered. Furthermore, you missed ABCAB in the two pairs cases.
[spoiler=How to compute it]
I use combinations: nCp = n! / ( p! (n-p)! ) , that give the number of different choices of p elements from a set of n elements when the order is not relevant.
And the factorial n! = n * (n-1) * (n-2) * ... * 3 * 2 * 1 .
So, the probability is the product of:
4C3 = 4 -> The number of sets of 3 suits we can have in our hand
13C1 = 13 -> The number of different values we can have in the trio
4C1 = 4 -> The number of sets of 1 suits we can have in our hand
4C1 = 4 -> The number of sets of 1 suits we can have in our hand (this term appears twice)
12C2 = 66 -> The number of sets of 2 values we can have in our hand
And this divided by
52C5 = 2 598 960 -> The number of sets of 5 cards we can have in our hand (from 52 cards in a deck).
So, the probability p is given by
p = 4C3 * 13C1 * 4C1 * 4C1 * 12C2 / 52C5
p = 4 * 13 * 4 * 4 * 66 / 2 598 960
p ≈ 0.02113
[/spoiler]
Regarding the line, you're right, I meant curve (like the Peano curve (http://en.wikipedia.org/wiki/Space-filling_curve), a fractal (http://en.wikipedia.org/wiki/Fractal)).
And about algebra... weren't you called a long time ago Algebra15 and latter (after your PC was stricken by lightning) Algebra17?
Yes, I was. And I did figure out that I missed one. And also, in poker, you're allowed to count higher hands as lower hands (if only for the sake of rubbing it in; I personally like to say, when everyone else has a poor hand, and I have 4 of a kind, "two pair -- jacks and jacks" or something like that).
What is the mathematical notation for the square route of negative 1? (-1)
The square root? That would be
The square route would be a two-block journey to 1 + 0i.
Osirus dont take this the wrong way but why did you start this tread?
Also I would just like to know what is in store for me. Is there any way you can define the difficulty of calculus to me.
First answer: Because I love mathematics, and some people have pmm'd me about mathematics, and I figured a thread would be easier.
Second answer: I'm no help to you there. I could tell you that calculus makes a lot of sense, it's perfectly intuitive, and you ought worry about nothing. However, that's just me; some people have an awful time trying to do calculus, and end up banging their heads against the walls. You may be one who believes calculus to be of the former, or one who experiences the latter in your time studying calculus.
As I do think math makes plenty sense of the real world than I guess I have nothing to worry about and just should accept the fact that I will have no (real) idea what it really must be like studying its field.