mm-448
Subject: Re: Largest Useful Number?Repeated for Xposting
purposes...in> the context of the brain (and afaics neural
connections). Is this a record> for a 'useful' number? Of
course it depends on how you define useful. Iwas> thinking in
terms of, say, upper bounds for proofs, the number
ofreferential> number y = x+1 used solely for saying y is 1
bigger than the number x your> rival found useful).> (btw
http://tinyurl.com/nb7s for another man obsessed with size)>
cheers> dd
===
Subject: Re: Largest Useful Number?> Repeated for
Xposting purposes...> in>> the context of the brain (and afaics
neural connections). Is this a record>> for a 'useful' number?
Of course it depends on how you define useful. I> was>>
thinking in terms of, say, upper bounds for proofs, the number
of> referential>> number y = x+1 used solely for saying y is 1
bigger than the number x your>> rival found useful).>> (btw
http://tinyurl.com/nb7s for another man obsessed with size)>>
cheers>> ddCheck out 's number
at<http://mathworld.wolfram.com/sNumber.html>. This number is
soenormous that a special notation had to be devised just to
express it.Not only does this number have a name, it is the
smallest known upperbound for the solution of a certain
problem in an area of graph theoryknown as Ramsey Theory.And
the kicker is: mathematicians suspect that the actual least
upperbound for the stated problem is 6, but they can't prove
it.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia
Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228
===
Subject: Re: Largest Useful
Number?<snip
===
Subject: Re: Largest Useful Number?> And the
kicker is: mathematicians suspect that the actual least upper>
bound for the stated problem is 6, but they can't prove it.I
notice that the page I cited says the number N* is now known
to be at least11 and is suspected to be larger.-- Dave
SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228
===
Subject: Cesaro MeanI have tried this for a
while now, but can't solve it...How do you show that the
Cesaro Mean converges to L as n converges toinfinity. I have
tried to solve this by trying to split up ?-sub n - L intotwo
parts and use the triangle inequality, and basically i have to
provethat ?-sub n -> x sub n as n -> infinity, since i get abs
(?-sub n - x-subn) + abs (x-sub n-L)< E.For anyone who isn't
familiar or just forgot the meaning, the nth Cesaromean is
defined as?-sub n= 1/n(x1+....x-subn). The idea is that as n
increases, the mean willeventually approach L the same way
that x-sub n approaches L, because allthe terms will be close
to L, and the average hence also close to L.Miao Xu
===
Subject:
Re: Smallest number without a name>Similarly, there is no
lowest number whose shortest name contains>fewer than a
hundred letters.> I would dispute this (with more substituted
for fewer, as you> acknowledge). The lowest number whose
shortest name contains more than a> hundred letters is not the
name of a number (and hence the source of a> paradox), it's a
*description* of a number. agreedbut how about the reciprocal
, or additive inverse (depending what youmean by small) of the
largest number ever namedI realize it is the same situation as
giving a person whose surname dependson some existing
unspecified event, but whose existence can be verified.For
example Whatever-the surname-of the -current-president-of-
the-U.S)This seems to be a peculiar Name and yet a Meta-name
(description of thealgorithm for determining a name) all at
once!!RJ Pease
===
Subject: Frustrated with math!! Please Help!I
know this problem should be very easy to work, but I am just
drawingblanks. Please help if you can!I am trying to solve for
x. 3 4___ _ ___ = 5x+2 x-2Hillarie
===
Subject: Re: Frustrated
with math!! Please Help!> I know this problem should be very
easy to work, but I am just drawing> blanks. Please help if
you can!> I am trying to solve for x.> 3 4> ___ _ ___ = 5> x+2
x-2> HillarieFirst thing you want to do is multiply both sides
by a common denominator,(x+2)(x-2).So doing that
gives3(x-2)-4(x+2)=5(x^2-4) Distribute3x-6-4x-8=5x^2-20
Simplify-x-14=5x^2-20 Get 0 on one side5x^2+x-6=0
Factor(5x+6)(x-1)=05x+6=0 or x-1=0 by Zero Product
Propertyx=-6/5 or x=1The check is left to you.David
Moran
===
Subject: Interest calculation over 1 yearMy credit
card company is currently offering me a deal where I can make
a0%bce transfer for the life of the bce as long as I spend 50
permonth on the card. The 50 would be charged at 17.9% APR.If
I were to spend 50 per month over a year, how much interest
would becharged? The salesperson calculated it at 71p for the
first month if thebce was 50.I am thinking of taking out a bce
transfer of 3000 and if I did notspend at least 50 per month, I
would be charged a rate of 6.9%.How exactly is interest
calculated from aprs so I can work it out myselfin
future?
===
Subject: Re: Interest calculation over 1
yearalt.math.undergrad, G Hampton >My credit card company is
currently offering me a deal where I can make a>0%>bce
transfer for the life of the bce as long as I spend 50
per>month on the card. The 50 would be charged at 17.9%
APR.>If I were to spend 50 per month over a year, how much
interest would be>charged?None, if you pay it within the grace
period. -- Stan Brown, Oak Road Systems, Cortland County, New
York, USA http://OakRoadSystems.comAddress munging may or may
not reduce the spam you get; it surelyreduces the number of
useful answers you get.
http://www.cs.tut.fi/~jkorpela/usenet/laws.html
===
Subject: Re:
Interest calculation over 1 year>alt.math.undergrad, G Hampton
> >>My credit card company is currently offering me a deal
where I can make a>>0%>>bce transfer for the life of the bce
as long as I spend 50 per>>month on the card. The 50 would be
charged at 17.9% APR.>>If I were to spend 50 per month over a
year, how much interest would be>>charged?>> >None, if you pay
it within the grace period. > Nope. This is a teaser. You
transfer your 3000 shekels (which get charged no interest).
Then you charge 50 denarii a month and pay 60 shekels a month.
All 60 shekels get credited against the 3000 no-interest
shekels and your 50 denarii are charged 111.5% APR.The easiest
way to calculate the APR is to set up a spreadsheet. Assuming
you don't want to get really super-sophisticated, you keep two
running bces. In column 1, you start with the bce transferred,
and decrease it by the payment each month, until you get to 0.
In column 2, you start with 0, and increment by 1.5% + 50.75
(assuming you charge your 50 on the first day of the month) or
50 (last day) or 50.375 (middle) (and except that the first
month it's 50). Once the first column gets to 0, you then
decrement the second column by the payment amount * 1.01 (this
assumes the payment is made about 10 days into the ing cycle)
and stop adding on the 50 that you're required to charge to
keep the zero-interest on the transferred amount. (You do this
with an if statement.)Just a note, the payment might also be
split between paying the interest on the interest-charging
part of the account and paying the rest against the
non-interest-charging part. I guess this means that you have
to read your contract. Also, I've heard that there are cards
that charge interest based on the two-cycle average daily bce,
rather than just the last cycle. You just adjust the
calculations accordingly.Anyway, in order to calculate the
APR, you now create a third column, which is the actual cash
flow. For the first month, it's the amount transferred, for
the subsequent months (until the transferred bce gets to 0)
it's the amount charged minus the amount paid (this will be a
negative number, meaning that it's cash out the door), and for
months after the transferred bce gets to 0, it's just negative
the amount paid (which is really the amount charged minus the
amount paid, since you're not charging anything that you'll be
charged interest on). Then you use the IRR function to
calculate the IRR, which is the monthly rate. Multiply by 12
to get the APR (because it's the law, that's why). Paying
$91.87 to pay it off in 5 years, I get 9.1%.Well, borrowing
$3000 and charging $50 a month, and paying $68.14 to pay it
all off in 10 years, I get 11.9%. (I have to do it in $
because my keyboard doesn't have a pound symbol.) You can
calculate the payment by using the solver function.As a
further piece of trivia, the reason these programs are called
spreadsheets is that, once upon a time, when you wanted to do
this sort of calculation, you'd pull out your ledger paper and
fill in the boxes, and the calculations would be spread out all
over the desks of the banks of clerks who were performing the
calculations. Sometimes people would tape the sheets to each
other so that they could see the calculations all together.
Anyway, even though it's a graphic on a screen, it's still a
spreadsheet. (And even though Roomba does the actual work, I
still vacuum the house, because I turn it on.)Jon
Miller
===
Subject: Mathematics - Bachelor's Degree QuestionsI'm
considering going to get my bachelor's degree in
mathematics.Couple questions...1) Is there good money to be
made in this major? If so, where's it at?2) Is there any way
to be a mathematician and still be your own boss (ie.notwork
for anyone)?--Anthony
===
Subject: Re: Mathematics - Bachelor's
Degree Questions> I'm considering going to get my bachelor's
degree in mathematics.> Couple questions...> 1) Is there good
money to be made in this major? If so, where's it at?> 2) Is
there any way to be a mathematician and still be your own boss
(ie.> not> work for anyone)?> --> AnthonyThere are careers in
operations research, being an actuary, statistics,
andobviously teaching. You might also consider taking courses
in a related areathat you like that might recognize a lot of
math. courses (so there is someoverlap in course requirements)
such as Electrical Engineering, Physics, orEconomics. Maybe
even a double major.What I've found that math. was helpful to
me in two ways: 1. The disciplinedway of thinking that you can
apply to problems 2. You can hold your own whensomething
requiring math. comes up - from statistics to
optimization.
===
Subject: Re: Mathematics - Bachelor's Degree
Questions> 2. You can hold your own when>something requiring
math. comes up - from statistics to optimization.Above quote
is one of the most important results of studying Math for any
majorfield of study which would require math. Mathematics
gives you powerful toolsfor clarifying quantitative situations
and making decisions for suchsituations. Combine the Math major
with another field. Engineers, chemists,scientific researchers,
all need to be able to use mathematics.G C
===
Subject: Re:
Mathematics - Bachelor's Degree Questions> I'm considering
going to get my bachelor's degree in mathematics.> Couple
questions...> 1) Is there good money to be made in this major?
If so, where's it at?If you are choosing your major based on
the ability to make money, you should be prepared to hate your
job. On the other hand, you *can* make a living with it.> 2) Is
there any way to be a mathematician and still be your own boss
(ie.> not> work for anyone)?I have heard of mathematicians
doing contract work for cities, etc. Not necessarily a
*reliable* income, mind you. Or you could be one of the people
posting that you'll solve their homework problems.--

===
Subject: Re: Mathematics - Bachelor's Degree Questions> I'm
considering going to get my bachelor's degree in mathematics.>
Couple questions...> 1) Is there good money to be made in this
major? If so, where's it at?If money's your thing, major in
business. Math is not a get quick scheme. After you graduate
you can get a job as a programmer, but that's not a get quick
scheme eithher.> 2) Is there any way to be a mathematician and
still be your own boss (ie.> not> work for anyone)?Stephen
Wolfram? Then again he's a pretty smart guy, his book
notwithstanding.
===
Subject: Re: Mathematics - Bachelor's
Degree Questionsprinted them out to further dwell
on.--Anthony>I'm considering going to get my bachelor's degree
in mathematics.>Couple questions...>1) Is there good money to
be made in this major? If so, where's it at?> If money's your
thing, major in business. Math is not a get quick> scheme.
After you graduate you can get a job as a programmer, but>
that's not a get quick scheme eithher.>2) Is there any way to
be a mathematician and still be your own boss(ie.>not>work for
anyone)?> Stephen Wolfram? Then again he's a pretty smart guy,
his book> notwithstanding.
===
Subject: Re: infinity dimensional
vector space> Can you prove that the vector space of
real-valued function over a> differentiable manifold M is
infinity-dimensionalI'm sure this assignment was given to you
to test whether YOU can.That makes it kinda moot whether
anybody else out here can. No?
===
Subject: Uniqueness of
gcd(a,b) = as + btHi everyone,It is well known that gcd(a, b)
= as + bt for some integers a, b, s, t, witha > b > 0. What I
need to know is whether s and t are unique. Any
hints?Bernd
===
Subject: Re: Uniqueness of gcd(a,b) = as + bt>
Hi everyone,> It is well known that gcd(a, b) = as + bt for
some integers a, b, s, t,with> a > b > 0. What I need to know
is whether s and t are unique. Any hints?> BerndSuppose they
arent. Suppose p,q do the job too. Then;as + bt = gcd(a, b) =
ap + bqa(s-p) = b(q-t)So how about letting;s-p = b => p =
s-bq-t = a => q = a+tSo;gcd(a, b) = ap + bq = a(s-b) + b(a+t)
= as + btas required.I hope. (Getting practice in before term
starts...!)Craig
===
Subject: Re: Uniqueness of gcd(a,b) = as +
bt>Hi everyone,>It is well known that gcd(a, b) = as + bt for
some integers a, b, s, t,> with>a > b > 0. What I need to know
is whether s and t are unique. Any hints?>Bernd> Suppose they
arent. Suppose p,q do the job too. Then;> as + bt = gcd(a, b)
= ap + bq> a(s-p) = b(q-t)> So how about letting;> s-p = b =>
p = s-b> q-t = a => q = a+tThe argument is bogus. Why not let
s - p = q - t = 0. Then p = s, q = t andap + bq = as + bt as
desired.> So;> gcd(a, b) = ap + bq = a(s-b) + b(a+t) = as +
bt> as required.> I hope. (Getting practice in before term
starts...!)Bernd
===
Subject: Re: Uniqueness of gcd(a,b) = as +
btNever mind, I just found a counter-example:gcd(5, 3) = 1 =
5(2) + 3(-3) = 5(5) + 3(-8)Here is another question which I'd
hoped to solve using the uniqueness ofgcd(a, b) = as + bt
(which I can't anymore): Suppose a, b are positiveintegers
with gcd(a, b) = 1. Then there exists positive integers s and
t,with 0 < s < b, such that as - bt = , and in particular (as)
mod b = 1. Showthat s is unique.Any hints on this one?Bernd> Hi
everyone,> It is well known that gcd(a, b) = as + bt for some
integers a, b, s, t,with> a > b > 0. What I need to know is
whether s and t are unique. Any hints?> Bernd
===
Subject: Re:
Uniqueness of gcd(a,b) = as + bt Visiting Assistant Professor
at the University of Montana.>Never mind, I just found a
counter-example:>gcd(5, 3) = 1 = 5(2) + 3(-3) = 5(5) +
3(-8)>Here is another question which I'd hoped to solve using
the uniqueness of>gcd(a, b) = as + bt (which I can't anymore):
Suppose a, b are positive>integers with gcd(a, b) = 1. Then
there exists positive integers s and t,>with 0 < s < b, such
that as - bt = , and in particular (as) mod b = 1. Show>that s
is unique.>Any hints on this one?You must assume that b>1,
obviously.Consider the collection { s in Z: s>0 and there is t
in Z such that as+bt = 1}The first thing to do is verify that
this set is nonempty. We knowthat there exist integers s and t
such that as+bt=1, but we do notknow anything about the signs
of s and t. If s is positive, you are done. The set is
nonempty. If s<0, thenthere exists k>0 such that bk>-s. Then1
= as+bt = a(s+kb) + b(t-ak)and s+kb>0. Thus, the set is
nonempty.Since it is a nonempty set of positive integers,
there must be asmallest s such that s>0, and there exists t
such that as+bt=1.Is s>=b, then1 = as + bt = a(s-b) +
b(t+a)and s-b>=0; it cannot equal 0, because then you would
have b(t+a)=1,which implies that b=1 or b=-1, which is
impossible. Therefore, thecondition s>=b is impossible. Thus,
0<s<b.Now explain why if you had TWO values s, s', both
satisfying 0<s,s'<b,and such that there exist t, t' such
thatas+bt = as'+ bt' = 1,then s=s'. (Hint: consider
s'-s)
===
Subject: Re: Uniqueness of gcd(a,b) = as + bt>Never
mind, I just found a counter-example:>gcd(5, 3) = 1 = 5(2) +
3(-3) = 5(5) + 3(-8)>Here is another question which I'd hoped
to solve using the uniqueness of>gcd(a, b) = as + bt (which I
can't anymore): Suppose a, b are positive>integers with gcd(a,
b) = 1. Then there exists positive integers s and t,>with 0 < s
< b, such that as - bt = , and in particular (as) mod b =
1.Show>that s is unique.>Any hints on this one?> You must
assume that b>1, obviously.> Consider the collection> { s in
Z: s>0 and there is t in Z such that as+bt = 1}> The first
thing to do is verify that this set is nonempty. We know> that
there exist integers s and t such that as+bt=1, but we do not>
know anything about the signs of s and t.> If s is positive,
you are done. The set is nonempty. If s<0, then> there exists
k>0 such that bk>-s. Then> 1 = as+bt> = a(s+kb) + b(t-ak)> and
s+kb>0. Thus, the set is nonempty.> Since it is a nonempty set
of positive integers, there must be a> smallest s such that
s>0, and there exists t such that as+bt=1.> Is s>=b, then> 1 =
as + bt = a(s-b) + b(t+a)> and s-b>=0; it cannot equal 0,
because then you would have b(t+a)=1,> which implies that b=1
or b=-1, which is impossible. Therefore, the> condition s>=b
is impossible. Thus, 0<s<b.> Now explain why if you had TWO
values s, s', both satisfying 0<s,s'<b,> and such that there
exist t, t' such that> as+bt = as'+ bt' = 1,> then s=s'.
(Hint: consider s'-s)OK. Say s' - s > 0 so a(s' - s) + bt'' =
1 for some integer t''.as' = 1 - bt', as = 1 - bt, and soa(s'
- s) + bt'' = 1 - bt' + bt - 1 + bt'' = b(t'' - t' + t) = 1
implying b= -1 or 1 which is impossible.The same happens if we
consider s - s' > 0. Therefore, by process ofelimination, we
must have that s' - s = 0 so s' = s. Very
good.Bernd
===
Subject: Re: Uniqueness of gcd(a,b) = as + bt
Visiting Assistant Professor at the University of Montana.>>
Now explain why if you had TWO values s, s', both satisfying
0<s,s'<b,>> and such that there exist t, t' such that>> as+bt
= as'+ bt' = 1,>> then s=s'. (Hint: consider s'-s)>OK. Say s'
- s > 0 so a(s' - s) + bt'' = 1 for some integer t''.>as' = 1
- bt', as = 1 - bt, and so>a(s' - s) + bt'' = 1 - bt' + bt - 1
+ bt'' = b(t'' - t' + t) = 1 implying b>= -1 or 1 which is
impossible.>The same happens if we consider s - s' > 0.
Therefore, by process of>elimination, we must have that s' - s
= 0 so s' = s. Very good.Actually, I gave you a very bad hint.
Mea culpa. There is no reason toassume that there is a t''
such that a(s'-s) + bt'' = 1.Instead, note that b must divide
a(s'-s), since as' + bt' - (as + bt) = 0 so a(s'-s) =
b(t-t').But b is coprime to a, so b must divide s'-s. However,
0<=s'-s < b, sothe only possibility is for s'-s = 0, or for
s'=s.
===
Subject: The only stupid question is ...Does anyone
know where I can find a web page that lists calculus symbols?
Idon't need to take the whole class just yet. I just need to
decipher a fewformulas -- I think I've got sigma (sum) and
delta (change) but there are afew more.TIA
===
Subject: Re: The
only stupid question is ...> Does anyone know where I can find
a web page that lists calculus symbols?I> don't need to take
the whole class just yet. I just need to decipher afew>
formulas -- I think I've got sigma (sum) and delta (change)
but there area> few more.A short list can be found in the
alt.algebra.help FAQ, to includesuggestions on how you might
express such symbols in
newsgroupese:http://aah.ryan-usa.com/node20.html--
Darrell
===
Subject: Proof: symmetric group is not cyclicI have
two ways to prove symmetric group is not cyclic if n is more
than 2.1)Since symmetric group is not abelian (n>2),and cyclic
group is abelian,so, symmetric group is not cyclic group
(n>2)2)Since symmetric group is a permutation group of order
n!,and cyclic group only has order n,so, when n>2, symmetric
group is not cyclic.P.L.
===
Subject: Re: Proof: symmetric group
is not cyclic Visiting Assistant Professor at the University of
Montana.>I have two ways to prove symmetric group is not cyclic
if n is more than 2.>1)>Since symmetric group is not abelian
(n>2),>and cyclic group is abelian,>so, symmetric group is not
cyclic group (n>2)This one is correct (assuming you have proven
that S_n is nonabelianwhen n>2).>2)>Since symmetric group is a
permutation group of order n!,>and cyclic group only has order
n,>so, when n>2, symmetric group is not cyclic.This one is
hopelessly wrong. Why could the symmetric group not becyclic
of order n!?
===
Subject: Re: Proof: symmetric group is not
cyclicin message <1ju9b.70891$PD3.4740307@nnrp1.uunet.ca>:> I
have two ways to prove symmetric group is not cyclic if n is
more than> 1)> Since symmetric group is not abelian (n>2),>
and cyclic group is abelian,> so, symmetric group is not
cyclic group (n>2)Yes, that works, but how do you know the
symmetric group S_nisn't abelian for n>2 ? (Hint: You only
need to show it for S_3,since that group is a subgroup of
every S_n, n>2.)> 2)> Since symmetric group is a permutation
group of order n!,> and cyclic group only has order n,> so,
when n>2, symmetric group is not cyclic.No. There's a cyclic
group of ever order, including n!.-- Jim Heckman
===
Subject:
Map Colouring in RP2It's a long time since I did this so I may
have some of these detailswrong but here goes. Part of a course
we did at college wasgeneralisations of the 4 colour map
theorem. IIRC in the realprojective plane the fewest colours
required to colour a map so thatno two countries which share a
border are the same colour is six. Wewere shown that six is
enough and then we were set the problem offinding a map that
required six colours. I found a solution I liked. Iwondered if
there were resources on this problem on the web? Is
thereanywhere I can see maps in RP2 that require six
colours?cheersdd
===
Subject: Math questions?!If anyone has any
advice or tips I would greatly appreciate it!~Hillarie1) A
simply supported beam of length 20ft supports a
uniformlydistributed load of 1000 pounds per foot. The bending
moment M infoot-pounds X feet from one end of the beam is given
by M=500X(20-X). a) Determine any points on the beam where the
bending moment iszero. b) Determine the positions on the beam
where the bendingmoment isless than 40,000 foot-pounds.*for
a), I did this: 0=500X(20-X) 0=10,000X-500X 0=9,500X *for b),
I did this: M<(500(40))(20-40) M<20,000(-20) M<-400,0002) The
equation that gives the height, h, of an object thrown
upwardwith an initial velocity of 20 feet/second is
h=20t-16t^2with t the number of seconds. After how many
seconds will the heightbe 4 feet above the ground?*for this
one I solved it like a regular quadratic equation. 3) The
total number of dollars spent on recreation in the
UnitedStates from 1980 to 1988 can be approximated by the
model Spending= 116.289 + 9.5067t + .84145t^2where the
spending is measured in ions of dollars and the time
trepresents the calendar year with t=0 corresponding to 1980.
Use thismodel to predict the year when total recreational
spending will reach$300,000,000,000.*for this one I factored
it like a regular quadratic equation.
===
Subject: Re: Math
questions?!> 1) A simply supported beam of length 20ft
supports a uniformly> distributed load of 1000 pounds per
foot. The bending moment M in> foot-pounds X feet from one end
of the beam is given by M=500X(20-X).> a) Determine any points
on the beam where the bending moment is> zero.> b) Determine
the positions on the beam where the bendingmoment is> less
than 40,000 foot-pounds.> *for a), I did this: 0=500X(20-X) >
0=10,000X-500X > 0=9,500X That's not right, because 500X(20-X)
is 10000 X - 500 X^2> *for b), I did this: M<(500(40))(20-40)>
M<20,000(-20)> M<-400,000I am not sure why you did that; this
just shows that the bending moment at X=40 is -400000. To do
what you need to do, you should start fromM < 40000500 X (20 -
X) < 40000and proceed from there> 2) The equation that gives
the height, h, of an object thrown upward> with an initial
velocity of 20 feet/second is> h=20t-16t^2> with t the number
of seconds. After how many seconds will the height> be 4 feet
above the ground?> *for this one I solved it like a regular
quadratic equation. Sounds good; what did you get?> 3) The
total number of dollars spent on recreation in the United>
States from 1980 to 1988 can be approximated by the model>
Spending= 116.289 + 9.5067t + .84145t^2> where the spending is
measured in ions of dollars and the time t> represents the
calendar year with t=0 corresponding to 1980. Use this> model
to predict the year when total recreational spending will
reach> $300,000,000,000.> *for this one I factored it like a
regular quadratic equation.Sounds good; what did you
get?meeroh-- If this message helped you, consider buying an
itemfrom my wish list:
<http://web.meeroh.org/wishlist
===
Subject: Re: Math
questions?!> If anyone has any advice or tips I would greatly
appreciate it!> ~Hillarie> 1) A simply supported beam of
length 20ft supports a uniformly> distributed load of 1000
pounds per foot. The bending moment M in> foot-pounds X feet
from one end of the beam is given by M=500X(20-X).> a)
Determine any points on the beam where the bending moment is>
zero.> b) Determine the positions on the beam where the
bendingmoment is> less than 40,000 foot-pounds.> *for a), I
did this: 0=500X(20-X)> 0=10,000X-500X> 0=9,500XYou did not
need to distribute the 500x. By the zero product property,
youcan break this up into 500x=0 or 20-x=0. Therefore x=0 or
x=20> *for b), I did this: M<(500(40))(20-40)> M<20,000(-20)>
M<-400,000In this case, M=40000
so40000=500x(20-x)10000x-500x^2=40000-500x^2+10000x=40000x^2-
200x=-800x^2-200x+800=0Solve that equation for x> 2) The
equation that gives the height, h, of an object thrown upward>
with an initial velocity of 20 feet/second is> h=20t-16t^2>
with t the number of seconds. After how many seconds will the
height> be 4 feet above the ground?> *for this one I solved it
like a regular quadratic equation.In this case, h=4 so
4=20t-16t^216t^2-20t+4=04t^2-5t+1=0(4t-1)(t-1)=0t=1/4 sec or
t=1 sec> 3) The total number of dollars spent on recreation in
the United> States from 1980 to 1988 can be approximated by the
model> Spending= 116.289 + 9.5067t + .84145t^2> where the
spending is measured in ions of dollars and the time t>
represents the calendar year with t=0 corresponding to 1980.
Use this> model to predict the year when total recreational
spending will reach> $300,000,000,000.> *for this one I
factored it like a regular quadratic equation.Your spending
will be 300, so plug that in, get 0 on one side and then
solvefor x.I'd work all the problems out, but I'm tired and am
about to hit the sack :)David Moran