==== i'm having some trouble with the following question: > Prove that the ideal (x + y^2, y + x^2 + 2xy^2 + y^4) in C[x,y] is a maximal ideal. my initial thought was to use hilbert's nullestellensatz, but i'm not sure how to factor the second polynomial (having the y term makes it pretty tough!). if anyone has any suggestions, i'd appreciate it. thanks, jay by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA13Umr31921; ==== Lora Bush wants to buy a photocopier. The salesperson has the following information on 3 models. if all 3 are used, a specific job can be done in 50 minutes. If copier A operates for 20 minutes and copier B for 50 mininutes, one-half of the job is finished. If copier B operates for 30 minutes and copier C for 80 minutes, three-fifths of the job is done. Which is the fastest copier, and how long does it takes for this copier to finish the whole job? Torsten is absolutely correct. I'll baby-step through the problem. (Don't be insulted.) Let copier A complete the job in a minutes, copier B in b minutes, and copier C in c minutes. In one minute, A can complete 1/a of the job, B can complete 1/b of the job, and C can complete 1/c of the job. Together, in one minute, they can complete: 1/a + 1/b + 1/c = 1/50 (of the job). That's Equation #1. In 20 minutes, A can complete 20/a of the job. In 50 minutes, B can complete 50/b of the job. Together, they complete: 20/a + 50/b = 1/2 (of the job). That's Equation #2. In 30 minutes, B can complete 30/b of the job. In 80 minutes, C can complete 80/c of the job. Together, they complete: 30/b + 80/c = 3/5 (of the job). That's Equation #3. We have a system of three equations in three variables (a,b,c). #1: 1/a + 1/b + 1/c = 1/50 #2: 20/a + 50/b = 1/2 ---> 2/a + 5/b = 1/20 #3: 30/a + 80/c = 3/5 ---> 3/a + 8/c = 3/50 And we can solve this system WITHOUT clearing the denominators. Multiply #1 by 8, and subtract #3: 8/a + 8/b + 8/c = 8/50 3/a + 8/c = 3/50 ----------------------- #4: 5/a + 8/b = 5/50 = 1/10 Now, solve the 2-by-2 system: #2: 2/a + 5/b = 1/20 #4: 5/a + 8/c = 1/10 Multiply #2 by 8, multiply #4 by 5, and subtract: 16/a + 40/b = 8/20 = 4/10 25/a + 40/a = 5/10 We get: 9/a = 1/10 ---> a = 90 minutes. Substitute this into #2: 2/90 + 5/b = 1/20 ---> b = 180 minutes. And into #3: 3/90 + 8/c = 3/50 ---> c = 300 minutes. Therefore, copier A is fastest with a time of 90 minutes. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9VEpXK08955; ==== >Lora Bush wants to buy a photocopier. The salesperson has the >following information on 3 models. if all 3 are used, a specific job >can be done in 50 minutes. If copier A operates for 20 minutes and >copier B for 50 mininutes, one-half of the job is finished. If copier >B operates for 30 minutes and copier C for 80 minutes, three-fifths >of >the job is done. >Which is the pastest copier, and how long does it takes for this >copier to finish the whole job? the solution x_i of the linear system 50*x_1 + 50*x_2 + 50*x_3 = 1 20*x_1 + 50*x_2 = 1/2 30*x_2 + 80*x_3 = 3/5 will the fraction of the job photocopier i is able to copy in one minute, and so 1/x_i is the time it takes for photocopier i to do the whole job. Best wishes Torsten. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA1Jo8q26880; ==== I do not knwo how to go about integrating the following expressions with respect to x..Can anyone give me some ideas??? e^[sqrt(3x+9)] and 1/(sinx-2) Ron X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA248G425689; ==== 1) INT e^[sqrt(3x+9)] dx Let u = sqrt(3x + 9), then u^2 = 3x + 9 And x = (u^2 - 9)/3 ---> dx = (2/3)u du Substitute: (2/3)INT u e^u du This can be integrated by parts: (2/3)(u - 1)e^u + C Answer: (2/3)[sqrt{3x + 9) - 1]e^[sqrt{3x + 9)] + C 2) INT dx/(sin x - 2) I suggest the substitution: z = tan(x/2) The following statements come from a long series of steps. So if you've never seen it, you'll have to take my word for it. sin x = 2z/(1 + z^2), dx = 2 dz/(1 + z^2) Substituting, the integral simplifies to: - INT dz/(z^2 - z + 1) We can complete-the-square in the denominator and get: - INT dz/[(z - 1/2)^2 + 3/4] Now, we can let: u = z - 1/2 and the integral is of the arctangent form. Good luck! X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA248E325665; ==== why not try using the heinemann books - they have one for each unit and are ace ==== >why not try using the heinemann books - they have one for each unit >and are ace What question was that an answer to? -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Address munging may or may not reduce the spam you get; it surely reduces the number of useful answers you get. http://www.cs.tut.fi/~jkorpela/usenet/laws.html X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA2JcXp20885; ==== 1!+2!+3!+...+n!=? in easy form ==== > 1!+2!+3!+...+n!=? in easy form I doubt it. If an easy form were known, it should have been given at . David X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA3DAb322085; ==== I don't know about your problem, but it's close to another sum result that is interesting. Namely, 1*1!+ 2*2!+ 3*3!+...+ n*n! = (n+1)! -1 The sum on the left is the largest integer that you can write in the standard Cantor representation b1*1!+b2*2!+b3*3!+...+bn*n! = M where all the bi are integers and 0<=bi<=i -- the largest number without having to use the next bigger factorial (n+1)!. For any integer M <(n+1)! it is easy to write M in the form above (uniquely). In fact, you can use a top-down recursive algorithm which compares M to multiples of n!, i.e. looks at floor[M/n!]=bn, or you can use a bottom up algorithm where you divide M successively by 2,3,4,5... and pick off the bi as remainders to get M=q1*2+b1=[q2*3+b2]*2+b1=q3*4!+b3*3!+b2*2!+b1*1! etc. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA4FV8t01405; ==== Why can't the theory of general relativity and quantum theory be combined to give a single thory? And what is this string theory stuff? Hmm.... X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA65F6803077; ==== Well, General relitivity relies on the geometry of space-time to be flat...smooth without the presence of matter. Quantum mechanics on the other hand, creates this quantum foam that distorts space-time on very small scales. When the two theories are put together they just dont jive mathematically speaking. You get inconsistent mathematical results such as 5=6 wich is purely impossible. Now for string theory. That smoothes out the quantum foam and the equations don't yeild anomalies. The downside is that these strings are so tiny (less than what is called planck's length) that it is impossible to verify there existance experimentally. It should be exciting to see what it developes into though. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA5H46o11620; ==== I'm not sure whether one of these has been done yet, but I just couldn't resist. Presenting a quick diagnosis of our patient, Mr. Harris: NARCISSISTIC PERSONALITY DISORDER - Has a grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements) I'm already one of the most powerful men on the planet. I'm a discoverer. I'm an artist. I am a hunter. --------------------------------------------- - Is preoccupied with fantasies of unlimited success, power, brilliance, beauty, or ideal love I've been cheated out of tens of thousands of dollars... I'll pay you $250,000 US from any one math prize that I win that exceeds that amount. I can make you rich, if you aren't rich already. If you are rich, I'll make you powerful. --------------------------------------------- - Believes that he or she is special and unique and can only be understood by, or should associate with, other special or high-status people (or institutions) - Has a sense of entitlement, i.e., unreasonable expectations of especially favorable treatment or automatic compliance with his or her expectations What's wrong with you people? The math is in my favor, while you're deluded. F---ING super math discovery... F---ING ERROR that's over a hundred years old... Saying that my proof is wrong [is] causing me financial harm. I reserve the right to seek redress [and] may do so in a court of law. --------------------------------------------- - Is often envious of others or believes that others are envious of him or her I hate you Magidin. You jealous b-st-rd. You sick twisted b-st-rd. --------------------------------------------- PARANOID PERSONALITY DISORDER - Suspects, without sufficient basis, that others are exploiting, harming, or deceiving him or her I'm currently being wronged by mathematicians... You've been lied to... They just want you to believe false math. My job is to end their fraud. It's my job to chase them down, as I am a hunter. ...these posters are engaging in fraud. --------------------------------------------- - Is preoccupied with unjustified doubts about the loyalty or trustworthiness of friends or associates But you're so dumb!!! Why can't any of you just accept mathematics? Mathematicians don't accept mathematics. Mathematicians have been running away in fear. Mathematicians live in a fantasy world Mathematicians are clearly terrified Mathematicians are such babies. Mathematicians are pathetic liars. Mathematicians are disgustingly bad liars when caught. I'm telling you, these people are EVIL. ...quit the lying you dark evil people I see you as evil incarnate ---------------------------------------------- - Persistently bears grudges, i.e., is unforgiving of insults, injuries, or slights - Perceives attacks on his or her character or reputation that are not apparent to others and is quick to react angrily or to counterattack Arturo Magidin is a rather evil person Arturo Magidin clearly is running away from the math. Why do you listen to proven liars like Magidin? Magidin lied to you. Magidin was, as I figured, lying. Magidin stands against the discipline of mathematics. I think he's actually, really evil. you anti-mathematician. What is wrong with you Magidin? you f---ing, stupid dumb-ss. You are a stupid Magidin piece of sh-t. He's bad, he's evil, and I'm sick of his crap. Nora Baron is trying to refute algebra!!! Nora Baron is full of it. You're trash Nora Baron. how much math do you actually know Nora Baron? You lack character and are a base liar. You're a despicable human being. You're disgusting trash. --------------------------------------------- Observe the sharp contrast between the above quotes and the following quotes, which appeared just a short while earlier. I've recognized that pissing off mathematicians will NOT get me anywhere... there is no way I'll get anywhere pissing off mathematicians. Trying personal attacks against me is what is silly. please pay attention to things like personal attacks... ...assertions made without presentation of logic or math. My position is that mathematics *can* be discussed without need for lots of emotion or histrionics. --------------------------------------------- X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA65SdS03822; ==== I am currently teaching 5th grade right now, and want to make the subject matter more interesting. I am looking for websites that give examples of manipulatives that can be used for different types of problems (i.e. addition with regrouping). I would love to see the different manipulatives that can be used. ==== James ==== >Sorry to post such trivia, but I think I'm going to pluck my eyeballs >out if I don't see a neat solution to this. Can anyone tell me the quickest way to solve for X in a situation like >this: x = 10^x ;) Yeah I know, I must be an idiot. If you take the log (base 10) of both sides: log x = x log 10 = x log x / x =1 Try graphing that and see if it makes any sense. IMHO... MJR (REMOVE NoSPAM. from REPLY address to reply) Michael J. Reeves, AA, ASc E-Mail: michaeljreeves@comcast.net --------------------------------------------------------- I have no SPAM. I don't give a SPAM. I take no SPAM from anyone. I am NOT in the SPAM business!!! ==== > Hey guys, need some help with this question please! Prove that the following groups are Abelian. Are (R{0}, *) and (C{0}, *) Abelian, and why would this be relevant? > what are their orders? Are > they cyclic (i) {z e C : z^12 = 1} e = Element, C = complex numbers If w = exp((2*pi*i)/12), what is w^12? Now consider positive integer powers of w. Are these in group (i)? Is w^n a periodic function of n? If so, what is its period? What elements of (i), if any, cannot be expressed as a power of w? > (ii) {z e R : z^12 = 1} Which elements of (i) are real numbers? -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will find these people and we will bring them to justice. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA63B3D26321; ==== hi jimmy, did you try long division or synthetic division? the first one comes out with no remainer. With the second one, try arranging the divisor in terms of decreasing powers of u, then divide. As for using matrices, I don't know where that comes from...this is pretty simple using plain algebra. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9VEMOY06951; ==== Sorry forgot to mention binary operations acting on both the groups are multiplication ==== Its been a while since I'v been in a math course and I'm trying to write a research paper. How do I figure out what the percentage of a number is if I have the total number? For instance, if I know that in the year 2000, there were 125 homicides in Massachusetts, and the population of Massachusetts at that time was 6,349,097. How do I figure out the percentage of people that were murdered? In other words, how do I figure out what the percentage that 125 is out of 6, 349,097. I tried dividing the two numbers but I got a number x 10(-5 power). Any idea how I translate this to an actual percentage? -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== You need to first think of this as a fraction, which you have: 125 / 6 349 097 Then you need to convert it to a decimal form, which you have: 1.968783... e-5 This is scientific notation to avoid writing tons of zeroes... but converted to a standard decimal, it's: 0.00001968783... This is still your fraction... so now you need to pretend you had only 100 people and figure the same fraction of those 100 people: 0.00001968783... x 100 Which gives you: 0.001968783... of 100 or 0.001968783... % Hope this helps. And hope I didn't make a mistake along the way and then post it. heh heh BJ MacNevin > Its been a while since I'v been in a math course and I'm trying to write a > research paper. How do I figure out what the percentage of a number is if I > have the total number? For instance, if I know that in the year 2000, there were 125 homicides in > Massachusetts, and the population of Massachusetts at that time was > 6,349,097. How do I figure out the percentage of people that were murdered? In other words, how do I figure out what the percentage that 125 is out of > 6, 349,097. I tried dividing the two numbers but I got a number x 10(-5 > power). Any idea how I translate this to an actual percentage? > -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== >Domain: >y <= x <= 2y >0 <= x <= 2 Now i want this as a double integral where dx is inner and dy as outer. >And apparantly it yields: >[0,1] dy [y,2y] f(x,y) dx + [1,2] dy [y,2] f(x,y) dx >I dont understand why, why is the integral splitted in two parts ? i dont know ascii notation for integrals so [0,1] means integral symbol >with upper and lower value. ==== > Domain: > y <= x <= 2y > 0 <= x <= 2 I think you meant 0 <= y <= 2. > Now i want this as a double integral where dx is inner and dy as outer. > And apparantly it yields: > [0,1] dy [y,2y] f(x,y) dx + [1,2] dy [y,2] f(x,y) dx > I dont understand why, why is the integral splitted in two parts ? Nor do I. What was the rest of the problem? > i dont know ascii notation for integrals so [0,1] means integral symbol > with upper and lower value. What you have is int ( int ( f(x,y), x = y..2*y), y = 0..2) = ( int ( f(x,y), x = y..2*y), y = 0..1) + ( int ( f(x,y), x = y..2*y), y = 1..2). Which is surely true. _Why_ you want to rewrite that way I don't know. -- Paul Sperry Columbia, SC (USA) ==== ----- Original Message ----- I think you meant 0 <= y <= 2. y <= x <= 2y 0 <= x <= 2 Let me clarify my question. First, we note that the domain can be rewritten x/2 <= y <= x 0 <= x <= 2 Which yields the integral [0,2] dx [x/2,x] f(x,y) dy The question is now, REVERSE the integration order of the integral. i DONT understand how to do that. The answer in my textbook reads: [0,1] dy [y,2y] f(x,y) dx + [1,2] dy [y,2] f(x,y) dx my problem is, i don«t understand WHY the reverse order integral can be rewritten that way. ==== > ----- Original Message ----- > > Domain: > y <= x <= 2y > 0 <= x <= 2 I think you meant 0 <= y <= 2. > > y <= x <= 2y > 0 <= x <= 2 OK. It looked odd to me and typos do happen. ( I also misread your answer. ) > Let me clarify my question. First, we note that the domain can be rewritten > x/2 <= y <= x > 0 <= x <= 2 > Which yields the integral [0,2] dx [x/2,x] f(x,y) dy > > The question is now, REVERSE the integration order of the integral. i DONT > understand how to do that. The answer in my textbook reads: [0,1] dy [y,2y] > f(x,y) dx + [1,2] dy [y,2] f(x,y) dx > my problem is, i don«t understand WHY the reverse order integral can be > rewritten that way. I'll try again. If you graph the region, you'll see a triangle with two slanty sides and one side parallel to the y-axis. The total variation of y's for that region is 0 <= y <= 2. But note for 0 <= y <= 1, the x's vary from one slanty side to the other. That is, y <= x <= 2y. However, for 1 <= y <= 2, the x's vary from the y = x side to the x = 2 side. That is y <= x <= 2. Hence the limits on your dxdy integrals. The key point is that, for the dydx integral, for any x between 0 and 2, the y's always vary from x/2 to x. However, for the dxdy integral and y between 0 and 2, sometimes the x's vary from y to 2y and sometimes the x's vary from y to 2. So you need two integrals. Instead of one region given by x/2 <= y <= x and 0 <= x <= 2, you have _two_ regions. One is given by y <= x <= 2y and 0 <= y <= 1 and the other region is given by y <= x <= 2 and 1 <= y <= 2. -- Paul Sperry Columbia, SC (USA)