mm-119
=Im 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] isa maximal ideal.my
initial thought was to use hilberts nullestellensatz, but
im notsure how to factor the second polynomial (having the y
term makes itpretty tough!). if anyone has any suggestions,
approve@localhost) 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 thefollowing information on 3 models. if all
3 are used, a specic jobcan be done in 50 minutes. If copier
A operates for 20 minutes andcopier B for 50 mininutes,
one-half of the job is nished. If copierB operates for 30
minutes and copier C for 80 minutes, three-fths ofthe job is
done. Which is the fastest copier, and how long does it takes
for thiscopier to nish the whole job?Torsten is absolutely
correct.Ill baby-step through the problem. (Dont 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 ofthe 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 (ofthe job).Thats 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).Thats 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).Thats 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 =
3/50And we can solve this system WITHOUT clearing the
denominators.Multiply #1 by 8, and subtract #3:8/a + 8/b +
8/c = 8/503/a + 8/c = 3/50---#4: 5/a +
8/b = 5/50 = 1/10Now, solve the 2-by-2 system:#2: 2/a + 5/b =
1/20#4: 5/a + 8/c = 1/10Multiply #2 by 8, multiply #4 by 5,
and subtract:16/a + 40/b = 8/20 = 4/1025/a + 40/a = 5/10We
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
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 specic 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 nished. If copier>B operates for 30
minutes and copier C for 80 minutes, three-fths>of>the job
is done. >Which is the pastest copier, and how long does it
takes for this>copier to nish the whole job?the solution x_i
of the linear system50*x_1 + 50*x_2 + 50*x_3 = 1 20*x_1 +
50*x_2 = 1/2 30*x_2 + 80*x_3 = 3/5will 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
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 expressionswith respect
to x..Can anyone give me some
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)] dxLet u = sqrt(3x +
9), then u^2 = 3x + 9And x = (u^2 - 9)/3 ---> dx = (2/3)u
duSubstitute: (2/3)INT u e^u duThis can be integrated by
parts: (2/3)(u - 1)e^u + CAnswer: (2/3)[sqrt{3x + 9) -
1]e^[sqrt{3x + 9)] + C2) 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 youve never seen it, youll
have to take my word for it.sin x = 2z/(1 + z^2), dx = 2
dz/(1 + z^2)Substituting, the integral simplies to: - INT
dz/(z^2 - z + 1)We can complete-the-square in the
denominatorand get: - INT dz/[(z - 1/2)^2 + 3/4]Now, we can
let: u = z - 1/2and the integral is of the arctangent
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 unitand are
ace
why not try using the heinemann books - they have one
for each unit>and are aceWhat question was that an answer
to?-- Stan Brown, Oak Road Systems, Cortland County, New
York, USA http://OakRoadSystems.comAddress munging may or may
useful answers you get.
(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 formI doubt it. If an easy form
were known, it should have been given
at<http://www.research.att.com/projects/OEIS?Anum=A007489>.
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hA3DAb322085;
=I dont know
about your problem, but its close to another sum resultthat
is interesting. Namely,1*1!+ 2*2!+ 3*3!+...+ n*n! = (n+1)!
-1The sum on the left is the largest integer that you can
write in thestandard Cantor representation
b1*1!+b2*2!+b3*3!+...+bn*n! = M whereall the bi are integers
and 0<=bi<=i -- the largest number withouthaving 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 whichcompares M to
multiples of n!, i.e. looks at oor[M/n!]=bn, or youcan use a
bottom up algorithm where you divide M successively
by2,3,4,5... 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.
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hA4FV8t01405;
=Why cant the
theory of general relativity and quantum theory becombined to
give a single thory? And what is this string
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
beat...smooth without the presence of matter. Quantum
mechanics onthe other hand, creates this quantum foam that
distorts space-time onvery small scales. When the two
theories are put together they justdont jive mathematically
speaking. You get inconsistent mathematicalresults such as
5=6 wich is purely impossible. Now for string theory.That
smoothes out the quantum foam and the equations dont
yeildanomalies. The downside is that these strings are so
tiny (less thanwhat is called plancks length) that it is
impossible to verify thereexistance experimentally. It should
(from approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hA5H46o11620;
=Im not sure whether one of these has been
done yet, but I justcouldnt resist.Presenting a quick
diagnosis of our patient, Mr. Harris:NARCISSISTIC PERSONALITY
DISORDER- Has a grandiose sense of self-importance (e.g.,
exaggeratesachievements and talents, expects to be recognized
as superior withoutcommensurate achievements) Im already one
of the most powerful men on the planet.Im a discoverer. Im
an artist.I am a
hunter.- Is
preoccupied with fantasies of unlimited success,
power,brilliance, beauty, or ideal love Ive been cheated out
of tens of thousands of dollars...Ill pay you $250,000 US
from any one math prize that I win thatexceeds that amount.I
can make you rich, if you arent rich already.If you are
rich, Ill make you
powerful.-
Believes that he or she is special and unique and can only
beunderstood by, or should associate with, other special or
high-statuspeople (or institutions) - Has a sense of
entitlement, i.e., unreasonable expectations ofespecially
favorable treatment or automatic compliance with his or
herexpectations Whats wrong with you people?The math is in
my favor, while youre deluded.F---ING super math
discovery...F---ING ERROR thats over a hundred years
old...Saying that my proof is wrong [is] causing me nancial
harm.I reserve the right to seek redress [and] may do so in a
court oaw.- Is
often envious of others or believes that others are envious
ofhim or herI hate you Magidin.You jealous b-st-rd.You sick
twisted
b-st-rd.PARANOID
PERSONALITY DISORDER- Suspects, without sufcient basis, that
others are exploiting,harming, or deceiving him or her Im
currently being wronged by mathematicians...Youve been lied
to...They just want you to believe false math.My job is to
end their fraud.Its my job to chase them down, as I am a
hunter....these posters are engaging in
fraud.- Is
preoccupied with unjustied doubts about the loyalty
ortrustworthiness of friends or associates But youre so
dumb!!!Why cant any of you just accept
mathematics?Mathematicians dont accept
mathematics.Mathematicians have been running away in
fear.Mathematicians live in a fantasy worldMathematicians are
clearly terriedMathematicians are such babies.Mathematicians
are pathetic liars.Mathematicians are disgustingly bad liars
when caught.Im telling you, these people are EVIL....quit
the lying you dark evil peopleI 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 notapparent to others
and is quick to react angrily or to counterattack Arturo
Magidin is a rather evil personArturo 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 gured,
lying.Magidin stands against the discipline of mathematics.I
think hes 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.Hes bad, hes evil, and Im
sick of his crap.Nora Baron is trying to refute
algebra!!!Nora Baron is full of it.Youre trash Nora
Baron.how much math do you actually know Nora Baron?You lack
character and are a base liar.Youre a despicable human
being.Youre disgusting
trash.Observe
the sharp contrast between the above quotes and the
followingquotes, which appeared just a short while
earlier.Ive recognized that pissing off mathematicians will
NOT get meanywhere...there is no way Ill 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 forlots of emotion or
histrionics.
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 thesubject
matter more interesting. I am looking for websites that
giveexamples of manipulatives that can be used for different
types ofproblems (i.e. addition with regrouping). I would
love to see thedifferent manipulatives that can be
used.
going to pluck my eyeballs>out if I dont 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
address to reply)Michael J. Reeves, AA, AScE-Mail:
michaeljreeves@comcast.net
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 numbersIf 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.
SmithThe vast majority of Iraqis want to live in a peaceful,
free world.And we willfind these people and we will bring
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 rst one
comesout with no remainer. With the second one, try arranging
the divisorin terms of decreasing powers of u, then divide. As
for usingmatrices, I dont know where that comes from...this
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 groupsare multiplication
=Its been a
while since Iv been in a math course and Im trying to write
aresearch paper. How do I gure out what the percentage of a
number is if Ihave the total number?For instance, if I know
that in the year 2000, there were 125 homicides
inMassachusetts, and the population of Massachusetts at that
time was6,349,097. How do I gure out the percentage of
people that were murdered?In other words, how do I gure out
what the percentage that 125 is out of6, 349,097. I tried
dividing the two numbers but I got a number x 10(-5power).
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 rst think of this as a fraction, which you have:125
/ 6 349 097Then you need to convert it to a decimal form,
which you have:1.968783... e-5This is scientic notation to
avoid writing tons of zeroes... but convertedto a standard
decimal, its:0.00001968783...This is still your fraction...
so now you need to pretend you had only 100people and gure
the same fraction of those 100 people:0.00001968783... x
100Which gives you:0.001968783... of 100 or0.001968783...
%Hope this helps. And hope I didnt make a mistake along the
way and thenpost it. heh hehBJ MacNevin> Its been a while
since Iv been in a math course and Im trying to write a>
research paper. How do I gure out what the percentage of a
number is ifI> 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 gure out the percentage of
people that weremurdered?>> In other words, how do I gure
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 <= 2I 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 dont know.-- Paul
SperryColumbia, SC (USA)
= Original Message >> I
think you meant 0 <= y <= 2.y <= x <= 2y0 <= x <= 2Let me
clarify my question. First, we note that the domain can be
rewrittenx/2 <= y <= x0 <= x <= 2Which yields the integral
[0,2] dx [x/2,x] f(x,y) dyThe question is now, REVERSE the
integration order of the integral. i DONTunderstand 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) dxmy problem is, i dont understand
WHY the reverse order integral can berewritten that way.
 Original Message >> Domain:> y <= x
<= 2y> 0 <= x <= 2>> I think you meant 0 <= y <= 2. y <= x <= 2y> 0 <= x <= 2OK. It looked odd to me and typos
do happen. ( I also misread youranswer. )> 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 dont
understand WHY the reverse order integral can be> rewritten
that way.Ill try again. If you graph the region, youll see
a triangle with two slanty sidesand one side parallel to the
y-axis.The total variation of ys for that region is 0 <= y
<= 2. But note for0 <= y <= 1, the xs vary from one slanty
side to the other. That is,y <= x <= 2y. However, for 1 <= y
<= 2, the xs vary from the y = xside to the x = 2 side. That
is y <= x <= 2. Hence the limits on yourdxdy integrals. The
key point is that, for the dydx integral, for any x between 0
and2, the ys always vary from x/2 to x. However, for the dxdy
integraland y between 0 and 2, sometimes the xs vary from y
to 2y andsometimes the xs 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 theother region is given by y <= x <=
2 and 1 <= y <= 2.-- Paul SperryColumbia, SC (USA)