mm-1169
===
Subject: x^(y+1) > (x+1)^y
If you test it out you \[CapitalThorn]nd that x^(y+1) > (x+1)^y is
true for all x >=2 and for all real yÕs.
In other words increasing the exponent to which a number is
raised inceases the \[CapitalThorn]nal number more than increasing the
number being
raised.
The question is how can one prove this? IÕve tried many
methods
but have been unable to prove it with Calculus, logarithms,
or any
combinations of them using a system of equations. However,
IÕve been
unable to prove it. The fact that there is no inherent
relationship
between x and y is especially problematic.
All solutions, suggestions, or ideas are welcome.
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===
Subject: Re: x^(y+1) > (x+1)^y
> If you test it out you \[CapitalThorn]nd that x^(y+1) > (x+1)^y is
> true for all x >=2 and for all real yÕs.
No. 2^301 < 3^300.
What is the REAL question? Perhaps
for what x and y is
x^(y+1) > (x+1)^y ?
You can rearrange this use logs and algebraic manipulation to
get a
bound on y in terms of x. You have to be a bit careful about
whether
x>1 or x<1. Perhaps the problem was If x>=2, for what y is
x^(y+1) >
(x+1)^y ?
Unless I messed up my algebra, y has got to be pretty small.
--
Kevin Karplus karplus@soe.ucsc.edu
http://www.soe.ucsc.edu/~karplus
life member (LAB, Adventure Cycling, American Youth Hostels)
Effective Cycling Instructor #218-ck (lapsed)
Professor of Computer Engineering, University of California,
Santa Cruz
Undergraduate and Graduate Director, Bioinformatics
Af\[CapitalThorn]liations for identi\[CapitalThorn]cation only.
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===
Subject: Re: x^(y+1) > (x+1)^y
My comments/suggestions are between the rows of *Õs, below.
> If you test it out you \[CapitalThorn]nd that x^(y+1) > (x+1)^y is
> true for all x >=2 and for all real yÕs.
*******************************************************
Really? No one has time to test it out for all x >= 2 and for
all
real yÕs.
What xÕs and yÕs have you used so far to test \
it out?
This is a nice homework problem. What course did it come from?
HereÕs a hint: When dealing with *equations* which involve a
variable as
an exponent, itÕs often helpful to take the log of each \
side.
Another hint: ItÕs a pre-calc level problem.
--- Joe
*******************************************************
> In other words increasing the exponent to which a number is
> raised inceases the \[CapitalThorn]nal number more than increasing the
number being
> raised.
> The question is how can one prove this? IÕve tried many
methods
> but have been unable to prove it with Calculus, logarithms,
or any
> combinations of them using a system of equations. However,
IÕve been
> unable to prove it. The fact that there is no inherent
relationship
> between x and y is especially problematic.
> All solutions, suggestions, or ideas are welcome.
--
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Subject: Graphpatch designed for teachers and students.
Product to enhance
the learning experience
GraphPatch specializes in high quality self-adhesive graph
labels for
Math and Science. Our product promotes neatness and accuracy
while
saving the teacher and students time. Our graphs can be
instantly
placed anywhere a graph is called for. Please visit our
website at
www.graphpatch.com or email us at graphpatch@aol.com
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===
Subject: heÕd sell shoes
Rethink the Cool + the Shoe
phil knight had a dream. heÕd sell shoes. heÕd \
sell dreams.
heÕd get rich. heÕd use sweatshops if he had \
to.
then along came a new shoe. plain. simple. cheap. fair.
designed for only one thing: kicking philÕs ass.
the unswoosher
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
$$$$$$$$$
For years, Nike was the undisputed champion of logo culture,
its swoosh an instant symbol of global cool.
Today, Phil KnightÕs Nike is a fading empire, badly hurt by
years of brand damage as activists and culture jammers
fought back against mind marketing and dirty sweatshop labor.
Now a \[CapitalThorn]nal challenge. We take on Phil at his own game - and
win.
We turn the shoes we wear into a counterbranding game. The
swoosh
versus the anti-swoosh. Which side are you on?
Adbusters has been doing R&D for more than a year, and guess
what?
Making a shoe - a good shoe - isnÕt exactly rocket science.
With a network of supporters, weÕre getting ready to launch
the
blackSpot sneaker, the worldÕs \[CapitalThorn]rst grassroots \
anti-brand.
You can help launch the blackSpot revolution.
THE BIG QUESTION:
Is it possible to take Phil KnightÕs billion-dollar
marketing momentum and, in a quick judo-like move, slap
him onto the mat with the power of his own PR thrust?
OUR KICK-ASS MARKETING STRATEGY >>
http://blackspotsneaker.org/02/
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it............................preorders@blackspotsneaker.org
sell
it...........................wholesale@blackspotsneaker.org
invest in
it......................investors@blackspotsneaker.org
support
it........................donations@blackspotsneaker.org
join the
jam........................jammers@blackspotsneaker.org
Make a straight donation... itÕs a worthy cause
with the potential to set an historic precedent
that could be repeated in other industries and
usher in more grass roots version of capitalism
in which megacorps do not control every area of
our childrenÕs lives.
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-0
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
$$$$$$$$$
Agha Moammar AL-Suqami
She can simply live inside Ziad when the upper lemons reject
in the long
ceiling.
===
Subject: Math Facts interactive ßash card web page
Please try my new Math Facts web page designed to help
children or adults practice their addition, subtraction,
multiplication, and division facts. Math Facts requires a
fairly recent browser (e.g. IE 5+ or Netscape 6.2+) - it
will not work with Netscape 4.7.
The URL is:
http://home.indy.rr.com/lrobinson/mathfacts/mathfacts.html
Math Facts can be customized for speci\[CapitalThorn]c lessons and it
logs accuracy and speed (temporarily, and for one student).
The site has no banner ads or popups.
I think this site might be useful for teachers to recommend to
students for home practice in math facts.
This site can also be downloaded to your PC/Macintosh if
you prefer to use it while not connected to the Internet.
Instructions for download are given in the \
ŌhelpÕ page.
Please let me know if you \[CapitalThorn]nd it useful, or if you have
comments, suggestions, or bug reports.
Lynn Robinson
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===
Subject: Solving the stated problem not the problem you think
up
Recently, several questions in this group made me thinking of
our math
education in general.
Seems to me we put too much emphasis on using daily language
to
describe math problems and in the process lost the beauty of
precise
of math. As a result, the students sometimes are trying to
solve a
problem they think are asked.
bashing the examine designers about item 8 of this test. see
item 8.
http://www.mathforum.com/epigone/mathed-news/clarcorplee
how the question was wrong because the top angle of the
triagle in the
question has to be 45 degrees, so the numbers in the probelm
can not
be right, and so on and so forth.
Notice that, in the problem, it never mentioned that the top
vertex of
the triangle is at the center of the octagon. So there is
only one
right way to solve the problem.
For the professor from Rice, she was clearly trying to solve
a problem
that has never been asked.
All above are not so annoy to me. My problem with the whole
episode is
that all this time, nobody ever voiced any support for the
teachers
and the test designers. (I noticed this thingy way too late.)
It was
used by the media as a tool to bash the standard test and
show how it
is not fair. For godÕs sake, a test can not be fair, it is
designed to
favor those studied harder! For me, even a ßawed test is
better than
none at all.
Now, I feel better. :-)
katy
PS: can somebody post this to math-ed group? I donÕt have
access to
that group.
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===
Subject: Re: Solving the stated problem not the problem you
think up
: For godÕs sake, a test cannot be fair, it is designed to
: favor those studied harder! For me, even a ßawed test is
: better than none at all.
:
: Now, I feel better. :-)
And now I feed better too!
On the other hand, as you pointed out, the test in question
was NOT ßawed. The ßaw lay with educators (and editors)
who couldnÕt read the question! Perhaps more correctly,
they donÕt remember the \[CapitalThorn]rst lesson of every \
high school
geometry course IÕve ever taught or taken - the diagram is
never to scale. DonÕt try to measure the diagram! The only
way to get any kind of wrong answer was to ignore the given
numbers and try to measure the diagram in some way.
In defense of my profession, that Math Professor from Rice U.
is not a Math professor. Her doctorate is in Curriculum.
Of course, everyone makes mistakes, but it must be most
embarrasing to have your mistake published in a major
newspaper
and used as the basis of education for an entire state!
Robert
|)|/| || Burnaby South Secondary School
|| |orewood@olc.ubc.ca || Beautiful British Columbia
Mathematics & Computer Science || (Canada)
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===
Subject: Re: Solving the stated problem not the problem you
think up
> Recently, several questions in this group made me thinking
of our math
> education in general.
> Seems to me we put too much emphasis on using daily
language to
> describe math problems and in the process lost the beauty
of precise
> of math. As a result, the students sometimes are trying to
solve a
> problem they think are asked.
> bashing the examine designers about item 8 of this test. see
> item 8.
> http://www.mathforum.com/epigone/mathed-news/clarcorplee
> how the question was wrong because the top angle of the
triagle in the
> question has to be 45 degrees, so the numbers in the
probelm can not
> be right, and so on and so forth.
> Notice that, in the problem, it never mentioned that the
top vertex of
> the triangle is at the center of the octagon. So there is
only one
> right way to solve the problem.
> For the professor from Rice, she was clearly trying to
solve a problem
> that has never been asked.
> All above are not so annoy to me. My problem with the whole
episode is
> that all this time, nobody ever voiced any support for the
teachers
> and the test designers. (I noticed this thingy way too
late.) It was
> used by the media as a tool to bash the standard test and
show how it
> is not fair. For godÕs sake, a test can not be fair, it is
designed to
> favor those studied harder! For me, even a ßawed test is
better than
> none at all.
> Now, I feel better. :-)
> katy
> PS: can somebody post this to math-ed group? I donÕt have
access to
> that group.
This is a very good point. While there is a convention in
standard
test that if a point look like it is located at the center of
the
circle, thne it is. However diagrams are also often draw
offscale so
not to give the answer away. And in this case no where is it
stated
the vertex is at the center of the octagon, so a student may
make such
an assumption \[CapitalThorn]rst, but not consider it as absolute fact when
the
answer does not agree (the so called alternate answer should
be 25
after rounding, it is de\[CapitalThorn]ntely not 27.)
There is no question should not be used, but it is because it
is
confusing, not because it is wrong.
In fact I think very few of the 4640 students whose grade was
changed
to pass because the problem was regraded really failed
because they
got stuck with this problem.
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Subject: Solving the stated problem not the problem you think
up
For godÕs sake, a test can not be fair, it is designed to
favor those studied harder! For me, even a ßawed test is
better than
none at all.
Now, I feel better. :-)
katy
You may feel better, but I feel confused. Yes, a test that
favors
those who study harder IS fairer. But how is a ßawed test
better than
none at all? How smart is a student going to feel when asked
to answer
a question that cannot be answered because it was badly
designed? How
would studying harder make the student any better equipped to
answer a
question that has no answer? How can a ßawed test be fair?
Now, I too, feel better.
mattsmom
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Subject: Quadratic Equations
Can anyone give me a hand on this Algebra problem? I came up
with an
answer, but I do not think itÕs right.
The kinetic energy of a moving body is given by E=1/2mv^2,
where E is
the kinetic energy, m is the mass, and v is the velocity.
What is the velocity of a moving body whose mass is 5
kilograms and
whose kinetic energy is 250 newton meters.
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Subject: Re: Quadratic Equations
> Can anyone give me a hand on this Algebra problem? I came
up with an
> answer, but I do not think itÕs right.
> The kinetic energy of a moving body is given by E=1/2mv^2,
where E is
> the kinetic energy, m is the mass, and v is the velocity.
> What is the velocity of a moving body whose mass is 5
kilograms and
> whose kinetic energy is 250 newton meters.
Ahh. Physics, the stuff you either love to hate or hate to
love (well, if
itÕs calculus based anyway ).
Well, youÕre given the Kinetic Energy and mass so we can \
\[CapitalThorn]nd
velocity.
E=250 and m=5 so the equation is
250=1/2(5)v^2 Multiply both sides of the equation by 2
500=5v^2 Divide both sides by 5
100=v^2 Take square root of both sides
10,-10=v
Negative velocity doesnÕt make sense so the answer is v=10
m/s.
David Moran
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===
Subject: Re: Quadratic Equations
> Can anyone give me a hand on this Algebra problem? I came
up with an
> answer, but I do not think itÕs right.
> The kinetic energy of a moving body is given by E=1/2mv^2,
where E is
> the kinetic energy, m is the mass, and v is the velocity.
> What is the velocity of a moving body whose mass is 5
kilograms and
> whose kinetic energy is 250 newton meters.
Ahh. Physics, the stuff you either love to hate or hate to
love (well, if
itÕs calculus based anyway ).
Well, youÕre given the Kinetic Energy and mass so we can \
\[CapitalThorn]nd
velocity.
E=250 and m=5 so the equation is
250=1/2(5)v^2 Multiply both sides of the equation by 2
500=5v^2 Divide both sides by 5
100=v^2 Take square root of both sides
10,-10=v
Negative velocity doesnÕt make sense so the answer is v=10
m/s.
David Moran
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===
Subject: How did they come up with the fact that 360 degrees
are in a
circle
How did they come up with the fact that 360 degrees are in a
circle?
my math teacher wanted our class to \[CapitalThorn]nd out and i have
scowered the
internet to \[CapitalThorn]nd nothing, can someone help me or give me a
lead,
adamwenner@triad.rr.com
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Subject: Re: How did they come up with the fact that 360
degrees are in a
circle
> How did they come up with the fact that 360 degrees are in
a circle?
> my math teacher wanted our class to \[CapitalThorn]nd out and i have
scowered the
> internet to \[CapitalThorn]nd nothing, can someone help me or give me a
lead,
> adamwenner@triad.rr.com
ItÕs not a fact so much as a de\[CapitalThorn]nition. Some \
de\[CapitalThorn]nitons are
more
convenient than others, though.
Try these:
http://mathforum.org/library/drmath/view/52546.html
http://www.wonderquest.com/circle.htm
http://www.madsci.org/posts/archives/may97/861222488.Ph.r.html
http://www.usatoday.com/news/science/wonderquest/2002-06-21-
circle.htm
--
Dave Thuleen
Escondido, California
dthuleen@cts.com
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===
Subject: How did they come up with the fact that 360 degrees
are in a
circle
http://www.usatoday.com/news/science/wonderquest/2002-06-21-
circle.htm
Or how about this one:
http://mathforum.org/library/drmath/view/59075.html
Kirby
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Subject: remainder theorem
When x^3 + cx + d is divided by x + 1, the remainder is 3,
and when it
is divided by x - 2, the remainder is -3. Determine the
values of c
and d.
I can get it down to c - d = 8 for the 1st part and (-2) - d
= 11 for
the 2nd part.
I think i subtract the 1st part by the 2nd part to get the
answer for
d then use that to get c but iÕm not sure.
answer is c = -14/3, d = -5/3
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Subject: Re:Remainder Theorem
This is what I get:
(1): x^3 + cx + d = (x+1).p(x) + 3, and
(2): x^3 + cx + d = (x-2).q(x) - 3; so
(1): (-1)^3 + c(-1) + d = 3, and
(2): (2)^3 + c(2) + d = -3; or
(1): -1 - c + d = 3, and
(2): 8 + 2c + d = -3; so
(1): d = 4 + c, and (substituting for d):
(2): 8 + 2c + (4 + c) = -3, or:
(2): 3c +12 = -3, or:
(2): c = -5, and therefore d = -1.
Check:
x^3 - 5x - 1 = (x + 1).p(x) + 3,
let p(x) = x^2 -x -4, then:
x^3 - 5x - 1 = (x+1).(x^2 - x - 4) + 3, it works.
(Find p(x) by backwards long division - choose terms to
eliminate
all the x-terms when multiplied out.)
Chances are you have a book with an error in the answer
section.
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===
Subject: Re: remainder theorem
>When x^3 + cx + d is divided by x + 1, the remainder is 3,
and when it
>is divided by x - 2, the remainder is -3. Determine the
values of c
>and d.
Remainder Theorm:
f(x)=(x-c)q(x)+r,
at x=c, f(c)=r
Use synthetic division for the numeric terms, of -1 and +2
for the
linear
factors x+1 and x-2.
Do NOT forget to include (0)x^2 as part of the polynomial
when you set up
your synthetic divisions.
for x+1, you get remainder d-1-c=3
for x-2, you get remainderd+8+2c=-3
You there have two equations and two unknowns to \[CapitalThorn]nd. Do as
matrices, or
substitution with substitution for last equation of the two ,
whichever
method:
c=-5 and d=-1
Your polynomial becomes
x^3 - 5x - 1
G C
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===
Subject: math
My child is at a sixth grade level and he needs to know the
de\[CapitalThorn]nitions for- an abundant number, a perfect number, and a
near
perfect number, and the catagory that hte nimber 200 \[CapitalThorn]ts \
into.
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Subject: algebra
i want to have a copy on the reveiw of related literature
about
algebra in making my theses on math.
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Subject: Saxon School Districts?
I may have asked this question about a year or so ago.
However, it is common knowledge that school districts change
their textbook
adoptions all the time.
So, let me ask once again.
Do you know of any school districts in the United States
which utilize the
Saxon Math textbook series for the middle school grades ( 6 ,
7, and 8 )?
I would like to update my list.
Dennis
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Subject: Re: Saxon School Districts?
> I may have asked this question about a year or so ago.
> However, it is common knowledge that school districts
change their
textbook
> adoptions all the time.
> So, let me ask once again.
> Do you know of any school districts in the United States
which utilize
the
> Saxon Math textbook series for the middle school grades ( 6
, 7, and 8 )?
> I would like to update my list.
I donÕt know of a good way to fact-check this, but \
IÕm pretty
sure the
Houston, Minnesota school district uses Saxon Math in all
grades, and
operates a charter school that enrolls pupils by distance
learning from all
over Minnesota. The charter school is part of the K-12
network of schools,
which I think prefers Saxon.
Hope this helps!
--
Karl M. Bunday Christ has set us free. Galatians 5:1
Learn in Freedom (TM) http://learninfreedom.org/
kmbunday AT earthlink DOT net (preferred email address)
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Subject: GraphSight lets your formulas talk (freeware
plotting utility
available)
IÕm glad to announce GraphSight v.2.0 - the curve \
\[CapitalThorn]tting and
math
graphing Windows software for scientists, teachers and
students.
Here is what GraphSight really does for you:
* Lets your formulas talk showing to you what they really
hide in
them.
* Helps you to get prepared to math lessons in time.
* Enables you to predict. It analyzes data and shows to you
the trend
and statistics.
* Simpli\[CapitalThorn]es your algebra, trigonometry or calculus homework.
* Prompts to you an optimal way of carrying out the solution
of
equation by visualizing it.
GraphSight is a convenient graphing and curve \[CapitalThorn]tting utility
for
Windows designed to quickly plot and explore 2D math
functions and
data. GraphSight easily plots common x(y), y(x) Cartesian,
parametric,
polar and table-de\[CapitalThorn]ned graphs. GraphSight curve \
\[CapitalThorn]tting engine
uses
least squares method to perform regression analysis (different
regression types are supported). This tiny powerful graphing
software
gives you a full control over a coordinate plane through the
interactivity of every graphics on it.
For more about GraphSight, please, navigate to:
http://www.cradle\[CapitalThorn]elds.com/products.html
--
Aliaksandr Murauski,
Cradle Fields - Software for Mathematics Study and Research
Work
http://www.cardle\[CapitalThorn]elds.com
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Subject: Basic Math Facts
learning add, sub, multi & divide for the past seven months.
I am
\[CapitalThorn]nally \[CapitalThorn]nished. I did this as a result of \
working as a
long-term
remedial math sub (Title I). I did not have materials I felt
I needed
and so I chose to develop what I wished I would have had. I
have been
a teacher for 40 years, but have been working as a sub the
past four
years since retirement. I found that what I was developing
worked
great in third grade also. If you are still interested, get
in touch
with me. My materials are not long, but they are structured
in such
a way that students must learn one portion of work, and prove
instant
recall, before they can go on. This avoids getting through
the 9Õs,
for example, and not knowing the 7Õs and 8Õs. \
My e-mail is not
working well for some reason, if you donÕt get through, \
write
me
if you are still interested in a new method.
Joan Sveen
1118 West Division St.
River Falls, WI 54022
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Subject: help
How would I solve the following:
^5
sqrt(125/0.55)
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Subject: Re: help
Type this in words please, your formatting got messed up.
What I think you want is the square root of (125/.55) raised
to the 5th
power.
This isnÕt a bad problem to solve.... just remember that \
when
taking the
square root of something, it is the same as raising that
something to the
(1/2) power. So if you have (blah) raised to the (1/2) power
and then that
raised to the 5th power, all you have to do is multiply the
exponents. So
(blah) ^ [(1/2)*(5)]. Also remember your ORDER OF OPERATIONS,
Please
If this wasnÕt your problem please repost with correct
problem.
Good Luck,
John
> How would I solve the following:
> ^5
> sqrt(125/0.55)
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Subject: Standard Form of Equation
How would I \[CapitalThorn]nd the standard form of equation (Ay+Bx=C) for a
line
passing (1,2) and (5,4)?
Chris
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