mm-491
===
Subject: Re: Surds
I worked out how to do it :)
just multiply everything by sqrt(5) and sqrt(3) so its 5*3 :)
and the same for the other
and for c) i worked out its 1/(sqrt(3)+1) = 1/(sqrt(3)+1) *
(sqrt(3)-1/(sqrt(3)-1 = (sqrt(3)-1)/2
I achieved something
Nathan
> Hi could someone please help me with these problems?
> a) 3/sqrt(5) + 1/sqrt(3) =
> b) 3 - 5/sqrt(3) =
> c) 1/(sqrt(5)+1) =
> d) 2/(2-sqrt(3) =
 Nathan

> Hints:
> a) Multiply first term by sqrt(5)/sqrt(5) (which =1), and
second term
by
> sqrt(3)/sqrt(3) (also =1)
> b) Multiply second term by sqrt(??)/sqrt(??) - you can
work out what ??
is.
> c) Multiply by (sqrt(5)-1)/(sqrt(5)-1)
> d) Use same sort of trick as for c) - but note its a minus
not a plus
in
the
> denominator ...
> Also. d) is missing a right-parenthesis. It should be
2/(2-sqrt(3)).
===
Subject: Re: Quadratic Equations - Problem Solving
hi
i just done
1a) and got x=-7 or 0 by using factorisiation (they were the 2
only
possible
answers that could work)
> Hi
> Could someone please help me with these quadratic equations?
> I must use the quadratic equation:
> 1a) The sqare of a number is equal to 7 times the number.
Which are the
two
> possible numbers?
> b) The length of this rectangle is 6m longer than the width.
If the
area
> is 40 m^2 find the value of x
> c) Find 2 consecutive positive numbers whos product is 30
> 2a) The sum of the first n positive integers is given by the
formula
> s=(n/2)(n+1) Find the number of intergers whose sum is 66
> i managed to get 2 (11 and -12) but this was just by guess
and check
> b)Find two numbers such that when 5 is added to the square
of the
number,
> the result is equal to 6 times the number
> Nathan
===
Subject: Re: Quadratic Equations - Problem Solving
hi
i will check over my worksheet over the weekend i have a major
geography
exam on this week and a history esssay to prepare
i will get back to yous
nathan :)
> Hi
> Could someone please help me with these quadratic equations?
> I must use the quadratic equation:
> 1a) The sqare of a number is equal to 7 times the number.
Which are the
two
> possible numbers?
> b) The length of this rectangle is 6m longer than the width.
If the
area
> is 40 m^2 find the value of x
> c) Find 2 consecutive positive numbers whos product is 30
> 2a) The sum of the first n positive integers is given by the
formula
> s=(n/2)(n+1) Find the number of intergers whose sum is 66
> i managed to get 2 (11 and -12) but this was just by guess
and check
> b)Find two numbers such that when 5 is added to the square
of the
number,
> the result is equal to 6 times the number
> Nathan
===
Subject: Re: Quadratic Equations - Problem Solving
> Hi
> Could someone please help me with these quadratic equations?
> I must use the quadratic equation:
> 1a) The sqare of a number is equal to 7 times the number.
Which are the
two
> possible numbers?
> b) The length of this rectangle is 6m longer than the width.
If the
area
> is 40 m^2 find the value of x
> c) Find 2 consecutive positive numbers whos product is 30
> 2a) The sum of the first n positive integers is given by the
formula
> s=(n/2)(n+1) Find the number of intergers whose sum is 66
> i managed to get 2 (11 and -12) but this was just by guess
and check
> b)Find two numbers such that when 5 is added to the square
of the
number,
> the result is equal to 6 times the number
> Nathan
1(a). Let x be the number(s) -- now translate into mathematical
notation:
The square of a number is equal to 7 times the number.
x^2 = 7x
So we need to solve the equation x^2 = 7x
You shouldn't need to use the quadratic formula to solve this;
instead,
(1) isolate zero, then (2) factor:
Isolate zero x^2 - 7x = 0 (subtract 7x from both sides of
the
original equation)
Now factor x(x - 7) = 0.
x and x - 7 are numbers -- for their product to be zero, one
or the other
of them must be zero:
x = 0 OR x - 7 = 0
from which
x = 0 OR x = 7 are the two numbers.
I'll leave it to you to verify that 0 and 7 each satisfy the
stated
conditions.
I hope you find this helpful.
Kevin O'Neill
===
Subject: Re: Quadratic Equations - Problem Solving
hi peter
please dont feel you are doing my homework for me,
i am just having troubles in maths but today i got the hang of
the surd
stuff :) espec like the ones like 1/(sqrt(3)+1) pretty good
stuff
but i still dont know how to do these equations. I can guess
and check them
and get the correct answwer but have no idea how to do them
a teacher said i could try using -b+-sqrt(..... the quadratic
forumula)
could you please porvide a few detailed answers to 1 or 2 of
the questions
.... that would be great
once again please dont feel that you are doing my homework for
me
> Doing your homework for you won't help you. I am sure your
textbook has
some
> worked examples. I am also sure your teacher would have
shown you some
> examples.
> I will help you - on this and on the one on surds - if you
put some
effort
> in yourself.
> First task.
> Write out each of the problems as an equation. I will do
1b), the
hardest,
> to show you what I want.
> Let L=length, W=width.
> L=W+6 (Length is 6 more than width)
> 40 = L*W (the area is 40, which in a rectangle is the length
times the
> width)
> 40 = (W+6)*W
> W*W + 6*W - 40 = 0
> There, thats the quadratic equation you have to solve. Now
do the same
for
> the others.
> Reply to this group with the equations for each ofd the
problems below.
> Hi
> Could someone please help me with these quadratic
equations?
> I must use the quadratic equation:
> 1a) The sqare of a number is equal to 7 times the number.
Which are the
> two
> possible numbers?
> b) The length of this rectangle is 6m longer than the
width. If the
area
> is 40 m^2 find the value of x
> c) Find 2 consecutive positive numbers whos product is 30
> 2a) The sum of the first n positive integers is given by
the formula
> s=(n/2)(n+1) Find the number of intergers whose sum is 66
> i managed to get 2 (11 and -12) but this was just by guess
and check
> b)Find two numbers such that when 5 is added to the square
of the
> number,
> the result is equal to 6 times the number
through
> Nathan
===
Subject: Re: Quadratic Equations - Problem Solving
As I said, you first have to write out each problem as an
equation.
I did 1b for you below.
At the bottom, I will do 1a.
> First task.
> Write out each of the problems as an equation. I will do
1b), the
hardest,
> to show you what I want.
> Let L=length, W=width.
> L=W+6 (Length is 6 more than width)
> 40 = L*W (the area is 40, which in a rectangle is the
length times the
> width)
> 40 = (W+6)*W
> W*W + 6*W - 40 = 0
> There, thats the quadratic equation you have to solve. Now
do the same
for
> the others.
> 1a) The sqare of a number is equal to 7 times the
number. Which are
the
> two
> possible numbers?
x^2 = 7x (x^2 means x squared)
This ones easy. One possible answer is x=0; the other one is
even more
obvious. You don't need the quadratic formula for that one
(though it will
work if you try it ... a=1, b=-7, c=0).
Now WRITE OUT THE OTHER PROBLEMS AS EQUATIONS. You don't have
to solve
them - I will help you do that if you can't - but you do have
to at least
try to write them as equations.
===
Subject: Re: Quadratic Equations - Problem Solving
hi peter
i done the width question and factorised ---> w*w + 6*w - 40
ti (x+10)(x-4)
and therfore x=-10 or x=+4 is this correct? i dont have answers
nathan
> As I said, you first have to write out each problem as an
equation.
> I did 1b for you below.
> At the bottom, I will do 1a.
> First task.
 Write out each of the problems as an equation. I will do
1b), the
> hardest,
> to show you what I want.
 Let L=length, W=width.
 L=W+6 (Length is 6 more than width)
 40 = L*W (the area is 40, which in a rectangle is the
length times
the
> width)
 40 = (W+6)*W
 W*W + 6*W - 40 = 0
 There, thats the quadratic equation you have to solve.
Now do the
same
> for
> the others.
> 1a) The sqare of a number is equal to 7 times the
number. Which are
> the
> two
> possible numbers?
> x^2 = 7x (x^2 means x squared)
> This ones easy. One possible answer is x=0; the other one is
even more
> obvious. You don't need the quadratic formula for that one
(though it
will
> work if you try it ... a=1, b=-7, c=0).
> Now WRITE OUT THE OTHER PROBLEMS AS EQUATIONS. You don't
have to solve
> them - I will help you do that if you can't - but you do
have to at least
> try to write them as equations.
===
Subject: Re: Quadratic Equations - Problem Solving
> hi peter
> i done the width question and factorised ---> w*w + 6*w - 40
ti
(x+10)(x-4)
> and therfore x=-10 or x=+4 is this correct? i dont have
answers
You don't need the answers to check it.
You can't have a minus number as a width, so that answer
doesn't apply.
For the other one, plug it back into the problem.
If the width is 4, the length is 6 plus the width, so its what?
This makes the are what?
So does this answer satisfy the questions?
===
Subject: Re: Quadratic Equations - Problem Solving
Hello
Yes it does work
Why cant I use a minus? It still works...
X*(x+6)=40
-10*(-10+6) = 40
4*(4+6)=40
Nathan
> hi peter
> i done the width question and factorised ---> w*w + 6*w -
40 ti
> (x+10)(x-4)
> and therfore x=-10 or x=+4 is this correct? i dont have
answers
> You don't need the answers to check it.
> You can't have a minus number as a width, so that answer
doesn't apply.
> For the other one, plug it back into the problem.
> If the width is 4, the length is 6 plus the width, so its
what?
> This makes the are what?
> So does this answer satisfy the questions?
===
Subject: Re: Quadratic Equations - Problem Solving
> Hello
> Yes it does work
> Why cant I use a minus? It still works...
> X*(x+6)=40
> -10*(-10+6) = 40
> 4*(4+6)=40
Because no real objects can have negative widths.
> Nathan
> hi peter
 i done the width question and factorised ---> w*w + 6*w
- 40 ti
> (x+10)(x-4)
> and therfore x=-10 or x=+4 is this correct? i dont have
answers
 You don't need the answers to check it.
> You can't have a minus number as a width, so that answer
doesn't apply.
> For the other one, plug it back into the problem.
> If the width is 4, the length is 6 plus the width, so its
what?
> This makes the are what?
> So does this answer satisfy the questions?
===
Subject: Re: Quadratic Equations - Problem Solving
> Hello
> Yes it does work
> Why cant I use a minus? It still works...
> X*(x+6)=40
> -10*(-10+6) = 40
> 4*(4+6)=40
> Nathan
No, it doesn't work, and it doesn't meet your conditions. And
its
important that you understand why.
Rectangles can't have lengths that are negative numbers. So
-10 is not a
valid length.
This happens a lot when you square numbers - you introduce a
new possible
solution that doesn't satisfy the original equation. For
example,
If x=6, the x^2 = 36.
But if x^2=36, then x can be +6 or -6.
So this happens a lot in quadratic equations, because they
always involve
squaring something. If the quadratic equation comes from
something in the
real world - like areas of rectangles, or calculating the
distance that
cannon balls fly, or calculating the energy to compress a
spring - then it
is very common that you will get two solutions, but one of
them is
physically impossible - because it involves negative lengths,
or firing
cannonballs from underground, or negative time. You will be
expected to
recognise when one of the solutions doesn't apply (because it
doesn't
satisfy the original question, even if it satisfies the
quadratic equation)
and eliminate it.
===
Subject: Dedekind cuts
Hi
Please give me a clear and simple outline as to what Dedekind
cuts are.
Karl
===
Subject: Re: Dedekind cuts
> Hi
> Please give me a clear and simple outline as to what
Dedekind cuts are.
> Karl
The clearest explanation I know is in the classic by G. H.
Hardy,
A Course of Pure Mathematics, 10th edition (1952 and reprints),
Chapter I.
Ken Pledger.
===
Subject: Re: Dedekind cuts
> Hi
> Please give me a clear and simple outline as to what
Dedekind cuts are.
> Karl
===
Subject: Re: New Problems
which is a good thing we have a test on surds and aglebra on
thurs which
counts to the school certificate. Anyway i was wondering if
you could help
me with a few of the following questions. These were the
harder questions
on
my sheet that i couldnt do . The questions are much easier
once i put them
into the quad. equation
B3) An n-sided polygon has (1/2n(n-3)) diagons. How many sides
has a figure
with 90 diagons?
B5) The was of this right-angles triangle is 3cm longer than
the height. If
the area is 20cm^2 find the value of x
C1) The difference between a number and 6 times its reciprocal
is 1. Find
the number
C3) Find the points of interesction of the curve y=x^2 and the
line
y=(2x+3)
c4) Te numerator of a fraction is 3 les than the denominator
of the
fraction. When the numerator is increased by 6 and the
denominator by 5,
the
value of the fraction is doubled. Find the fraction.
If you could help me with Any of these Questions that would be
great
NAthan
> Hi
> Could someone please help me with these quadratic equations?
> I must use the quadratic equation:
> 1a) The sqare of a number is equal to 7 times the number.
Which are the
two
> possible numbers?
> b) The length of this rectangle is 6m longer than the width.
If the
area
> is 40 m^2 find the value of x
> c) Find 2 consecutive positive numbers whos product is 30
> 2a) The sum of the first n positive integers is given by the
formula
> s=(n/2)(n+1) Find the number of intergers whose sum is 66
> i managed to get 2 (11 and -12) but this was just by guess
and check
> b)Find two numbers such that when 5 is added to the square
of the
number,
> the result is equal to 6 times the number
> Nathan
===
Subject: Re: New Problems
If you tried framing any of these as an equation I would
happily help.
> which is a good thing we have a test on surds and aglebra on
thurs which
> counts to the school certificate. Anyway i was wondering if
you could
help
> me with a few of the following questions. These were the
harder questions
on
> my sheet that i couldnt do . The questions are much easier
once i put
them
> into the quad. equation
> B3) An n-sided polygon has (1/2n(n-3)) diagons. How many
sides has a
figure
> with 90 diagons?
> B5) The was of this right-angles triangle is 3cm longer than
the height.
If
> the area is 20cm^2 find the value of x
> C1) The difference between a number and 6 times its
reciprocal is 1. Find
> the number
> C3) Find the points of interesction of the curve y=x^2 and
the line
y=(2x+3)
> c4) Te numerator of a fraction is 3 les than the denominator
of the
> fraction. When the numerator is increased by 6 and the
denominator by 5,
the
> value of the fraction is doubled. Find the fraction.
> If you could help me with Any of these Questions that would
be great
> NAthan
> Hi
> Could someone please help me with these quadratic
equations?
> I must use the quadratic equation:
> 1a) The sqare of a number is equal to 7 times the number.
Which are the
> two
> possible numbers?
> b) The length of this rectangle is 6m longer than the
width. If the
area
> is 40 m^2 find the value of x
> c) Find 2 consecutive positive numbers whos product is 30
> 2a) The sum of the first n positive integers is given by
the formula
> s=(n/2)(n+1) Find the number of intergers whose sum is 66
> i managed to get 2 (11 and -12) but this was just by guess
and check
> b)Find two numbers such that when 5 is added to the square
of the
> number,
> the result is equal to 6 times the number
through
> Nathan
===
Subject: Re: New Problems
whats framing?
> If you tried framing any of these as an equation I would
happily help.
now
> which is a good thing we have a test on surds and aglebra
on thurs
which
> counts to the school certificate. Anyway i was wondering
if you could
help
> me with a few of the following questions. These were the
harder
questions
> on
> my sheet that i couldnt do . The questions are much easier
once i put
them
> into the quad. equation
> B3) An n-sided polygon has (1/2n(n-3)) diagons. How many
sides has a
> figure
> with 90 diagons?
> B5) The was of this right-angles triangle is 3cm longer
than the
height.
> If
> the area is 20cm^2 find the value of x
> C1) The difference between a number and 6 times its
reciprocal is 1.
Find
> the number
> C3) Find the points of interesction of the curve y=x^2 and
the line
> y=(2x+3)
> c4) Te numerator of a fraction is 3 les than the
denominator of the
> fraction. When the numerator is increased by 6 and the
denominator by
5,
> the
> value of the fraction is doubled. Find the fraction.
> If you could help me with Any of these Questions that
would be great
> NAthan
> Hi
> Could someone please help me with these quadratic
equations?
> I must use the quadratic equation:
> 1a) The sqare of a number is equal to 7 times the
number. Which are
the
> two
> possible numbers?
> b) The length of this rectangle is 6m longer than the
width. If the
> area
> is 40 m^2 find the value of x
> c) Find 2 consecutive positive numbers whos product is 30
 2a) The sum of the first n positive integers is given by
the formula
> s=(n/2)(n+1) Find the number of intergers whose sum is 66
> i managed to get 2 (11 and -12) but this was just by
guess and check
> b)Find two numbers such that when 5 is added to the
square of the
> number,
> the result is equal to 6 times the number
 through
> Nathan

===
Subject: Order of Complexity
Hi all,
I am doing this course on algorithm analysis. I have problem
understanding
how does one see if an algorithm has complexity of n log n,
log n or log
log
n. I found this on the web:
If the computing time for mergesort is T(n) and that for merge
is cn, then:
T(1)=a=constant (the time it takes to call the routine with a
single
number,
do nothing and return), and
T(n)=2T(n/2)+cn, for n>1 (two calls to mergesort and one call
to merge)
If n is a power of 2, say n = 2k:
T(n) = 2T(n/2)+cn = 2(2T(n/4)+cn/2)+cn = 4T(n/4)+2cn =
4(2T(n/8)+cn/4)+2cn
=
8T(n/8)+3cn = ... = 2^kT(1)+kcn =an+cnlogn
If n is not a power of 2, it is less than some 2k, so T(n) is
bounded by
the
above. Therefore T(n)=O(nlogn) for all n.
How does one see that 2^kT(1)+kcn =an+cnlogn? Hope someone can
help me out
Thuan Seah