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#9826 Re: Help Me ! » provement » 2010-09-07 08:15:27

Bob

hi bobbym

How are you today?  Thanks for once again trying.

Bob

#9827 Re: Exercises » problem with probabilities » 2010-09-07 08:12:26

Bob

hi leealden06

I've met this before.  Are you asking how to calculate this?

Bob

#9828 Re: Help Me ! » Scientific Notation » 2010-09-07 08:08:06

Bob

Hi Dave

Your log fire reminded me of the following:

roflol

bob

#9829 Re: Help Me ! » equivalence relation questions » 2010-09-06 23:00:07

Bob

Thanks Avon,

That helps me too.

Back to Question 2.

The sets denoted Si and Sj don't have anything in their intersection so this partition divides main set S into non overlapping subsets.

Construct a relationship, R,  between elements that basically says a and b are related ( a R b) if and only if a and b are both in the same Si.

It's reflexive because a R a (a and a are both in the same set).

It's symmetric because if a R b this means they're both in the same Si which means b R a.

It's transitive because if a R b, and b R c this means that a, b and c are all in the same Si so a R c.

That makes it an equivalence relationship.

Bob

#9830 Re: Help Me ! » equivalence relation questions » 2010-09-06 07:20:29

Bob

hi nha,

Not my best subject but, as no one else is posting I'll throw in my ideas.

(ii) first.  I think you've got to show symmetric, reflexive and transitive. 

So, for example, define R so that a R b. when a and b are both in Si. 

As  a in Si and b in Si => b in Si and a in Si => b R a so it is reflexive.

(i)  ??? What ordered pairs is this question after?  I really cannot get this.  Sorry.  Do you have a given example already to show what this means?

Bob

#9831 Re: Help Me ! » Why does an Linear Congruential Generator work? » 2010-09-06 06:59:11

Bob

Hi George Y,

It must be 40 years since I did this so what follows is 'covered in mathematical rust'. 

Cannot recall that it is a named theorem at all.  I had a quick trawl on Wiki but found nothing.  But it may be there somewhere.  Best guess is Gauss discovered/proved it first.

To prove:

(i)  You won't get a zero remainder.

Assume you do.  => axn (for some integer n) = b x m (for some other integer m)

=> a and b have a prime factor other than 1 =><=

(ii)  Assume two integers give the same remainder

=> axn = axm for different integers n and m

=> a(n-m) = 0

=><= (i)

(i) and (ii) together => every remainder will occur once and once only.

Bob

#9832 Re: Help Me ! » provement » 2010-09-06 06:38:47

Bob

Hi bobbym,

Thank you for posting on this.  At least I can feel as though someone is 'out there'.  This is an excellent example of what we were saying earlier.

He posted new questions faster than I could respond.

What puzzles me is the spread of levels of difficulty.  From basic sine work through to a fairly complex circumcircle question which took me a page of working, using two advanced trig formulas followed by a binomial expansion.  I'm not claiming that's the only way to do the problem but, however you do it, it's several orders of magnitude harder than "why is sin 0 =0 ?"

I'd could help if I could get a 'feel' for what the course is and what level the questioner is at with the work.  But, without a response I cannot do more.  If you're reading this sameer, TALK TO US!.

Bob

#9833 Re: Help Me ! » provement » 2010-09-06 01:57:14

Bob

Aaaarrrhhhh.  Yet another question.  Please slow down and take it one at a time.  I need to know you that understand the two sine questions.  There's no point me going on to the range and circumcircle questions until I get your answer to that.

Latest question:  Do you mean ?

Also, looking back, I cannot find any response from you to posts 13, 14 , 17 and 18.  When you ask for help and someone posts you a question in order to help, you need to respond or we'll lose interest.


bob

#9834 Re: Help Me ! » provement » 2010-09-06 01:45:20

Bob

hi sameer,

Too many questions at once.  Just started to look at the range question.

But the sine ones are easier so I'll start there.

When you first meet sine and cosine it's in a right angled triangle.  diagram below

sine A = O/H

so when A = zero Opposite = zero, so sinA = zero

If you start with an equilateral triangle (all sides = 2) and cut it down the middle one half is a right angled triangle.

The angles are 30, 60 90 and H= 2, 'O' = 1 so sin 30 = 1/2.

The range question suggests to me that you are now onto the more general definition of sine cosine and tangent which allows any angle not just those in a right angled triangle.  (see second diagram)


More will follow .... just want to check what I've done so far.  Please reply so I know you're ok so far.

Bob

#9835 Re: Help Me ! » Permutation question » 2010-09-05 09:36:23

Bob

Thanks guys;  I should have got that! I feel such a fool.rolleyes:

Bob

#9836 Re: Help Me ! » Permutation question » 2010-09-05 02:03:40

Bob

Hi Carisma,

A committee of five people is to be chosen from 5 married couples. In how many ways can the committee be chosen if:

c. both husband and wife cannot be on the committee

I thought I could do these questions easily but I'm getting an answer in 3 figures (?).

Help me out here.  What answers did you get for a and b?

Bob

#9837 Re: Help Me ! » simplex splitting using center of gravity » 2010-09-04 20:47:22

Bob

hello vlnikolic

I'm going to try to make a model using Sketchup.  Then I'll need a big think so I may be gone some time.

Bob

#9838 Re: Help Me ! » provement » 2010-09-04 20:23:52

Bob

hi bobbym,

I agree.  I like helping people with their maths, but, if they're genuinely stuck on homework and wnat to learn more about the topic, then just a straight 'heres the answer' isn't going to do them any good.  They won't be able to make a post when they're doing their exam!

I'd prefer it if someone said: 'Its my homework and I want to get the answer myself but please could someone point me in the right direction.'

It would also be courteous to say thanks afterwards.  Or even some feedback that says whether the help provided actually clarified it for them.  If someone posted back and said 'Im still a bit confused about ......'.  No problem to add more help where it's genuinely needed.

For anyone else reading this, please take note!

Bob

ps.  bobbym, I'm still working on the Cauchy problem.  Thought I had it the other day, but I just went round in a perfect circle and landed back where I'd started.

pps  D'ya think they think we get paid for this?

#9839 Re: Help Me ! » simplex splitting using center of gravity » 2010-09-04 08:10:23

Bob

hi vlnikolic

I cannot be sure that 3D works this way.  My problem with that case is I cannot get a clear picture in my mind of the solids you are making.  And also, what do you mean by 'convex' for a 3D solid.  Have you ever used google sketchup.  It's a free download and then you could make your solids and rotate them.  And send me a copy of the file.  Let me know if you want to try this.

Bob

#9840 Re: Help Me ! » provement » 2010-09-04 07:51:29

Bob

Hi

There are too many threads muddled up here.  Can someone split them please?

There is a formula for working out the area of any triangle (doesn't require a right angle)

where a, b, c are the lengths of the sides and

But you need to know x first, unless what you are wanting is a formula for the area in terms of x.

sameer mishra: You need to respond to the questions you have been asked about both of the problems you have posted.

Bob

#9841 Re: Introductions » Hello everyone! I want to learn to teach math. » 2010-09-03 08:58:02

Bob

hi learn2teach

I don't know how the grades equate.  That year 10 class was a 'bottom set' ie weakest in the year at math.

Lower down the page you'll find SATs and GCSE papers with a higher level of question.

Bob

#9842 Re: Help Me ! » provement » 2010-09-03 08:38:54

Bob

hi again,

to prove tan20.tan30.tan40 = tan10

















thus putting (1) and (2) together

#9843 Re: Introductions » hello math lovers » 2010-09-03 04:31:47

Bob

hi everyone and elliot60 especially,

If it's easy there's no satisfaction in doing the problem ... and if it's too hard, then it's upsetting to be unable to do it.

So for me the optimum problem is hard (for me) but just at the limit of what I can do.  Is this a good philosophy for life?

Bob

#9844 Re: Introductions » Hello everyone! I want to learn to teach math. » 2010-09-03 04:14:59

Bob

hi learn2teach,

I think my response is unchanged.  Afterall, it's what you do that's important; not what the rest of the profession do!

So, to provide a good education, I would recommend you teach them to think.

To help them to pass the tests, proceed as follows.

(i) Get as many old test papers as you can to use as practice tests.

(ii) Give them a test and mark it.

(iii) Based on the 'failed' questions and areas that you identify as causing a problem, make a hints sheet that helps them towards an answer without actually giving the answer.

(iv)  Give the class a limited amount of time to do corrections using your hints.  Circulate to encourage and give additional help if the hint isn't enough.  TARGET: to improve their mark by 15%.

(v) Repeat the process with the next paper.

I have done this for years with SATs pupils (UK age 13/14) and GCSE students (UK age 15/16)

You will find examples of hints sheets on my website www.bundy.demon.co.uk/lessons

These are based on questions that you haven't seen of course but, it'll give you an idea what I mean.  Mostly they are WORD documents so you are welcome to download any you like and extract bits that might be useful to you.  The same goes for any others on the site.  I did these over many years; some are lesson plans and refer to specific text books so will be less useful; but others are there to provide help on topics that might be useful to others.

Now I'm retired, I've got time to sort out these pages and take off the stuff that's not really much help to make room for new stuff by request so any feedback would be welcome.

Bob

#9845 Re: Help Me ! » simplex splitting using center of gravity » 2010-09-03 03:47:30

Bob

Hi vlnikolic,

Triangle case.

I'm assuming a diagram like image two below;  ie. G lies at the intersection of the medians.

And you want to prove that image one cannot occur.

Consider the diagram in image three.

ABC is the triangle,   D, E and F are the midpoints of BC, AC and AB.

Join D to F and let it intersect BE at G.

BDF is similar to BCA (ie. same angles) and half the size, so BG = ½ BE.

Angle FGD = 180

For angle FGD to be reflex, G must be nearer to B than half way along BE.

But the centre of gravity, G, is only one third of the way up the median so this will never happen.

Bob

#9846 Re: Help Me ! » provement » 2010-09-03 03:04:07

Bob

hi

3x + 4y + 5z = 12 is the equation of a plain in 3D and the other is the square of the distance from the origin to the plain.  The minimum distance is the length of a line from the origin, that is perpendicular to the plain.

Do you know how to get the vector form for such a line ?

If not, I'll post it.

Bob

#9847 Re: Help Me ! » Help with Logs » 2010-09-03 01:12:12

Bob

Hi Natkirky

Start with the expression not yet squared.

split the powers:

take the '3' inside the log

but 'e' and 'ln' are inverse functions

now square the expression and re-order the terms:

Hope that helps, (but post back if any step is unclear)

Bob

#9848 Re: Help Me ! » Help with Logs » 2010-09-02 23:40:54

Bob

Hi Natkirky,

But where is the 'equals' in your equation ?

This may make it easier.

Bob

#9849 Re: Puzzles and Games » Complete the series.... » 2010-09-01 01:08:21

Bob

Hi bobbym,

How are you today?

Thanks for the hide help.  That sorts it out for me.

With regard to JaneFairfax's Liverpool question,  I think she's set a tough one if you're in the USA. 

Back in July you advised me  "R. I. P. O. S. T. P."

So what happened to English soccer in the year ?

 

Even so, you're in for a lot of gooling.

Bob

#9850 Re: Puzzles and Games » Complete the series.... » 2010-08-31 01:55:40

Bob

Hi JaneFairfax,

Thanks for the hide help.  But I've still not fully grasped it.  You can create a box which has a 'title' embedded in it. 

eg. 

but where do I put the hidden content?

Bob

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