Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

hi kapitanamerika

Welcome to the forum.

I suspect a catch question. You have 12 matches. I notice that you can make the four letters of NINE (a square number) using 11 matches. One snag: I need to move 4 matches not 3.

Someone will come up with a better solution I'm sure.

Bob

I made a drawing using 'Sketchpad', and got about 3.6

That's not to say I'm disagreeing with your answer. I may not have been sufficiently accurate. Please post your method.

Bob

hi Mystogan

That series is a geometric progression so you can use the formula:

http://www.mathsisfun.com/algebra/seque … etric.html

Bob

Apparently, but really there's only two: the 'in it for how much I can make' party and the 'I really want to make the world a better place' party, which is everyone else.

Unfortunately, the media tend to promote the former.

Bob

If the centre is point H, you can enter all the angles around H and hence get angle HDG in an isosceles triangle. So splitting it in half gives you a 90 angle for some trig.

Bob

hi lovealgebranotgeometry

You can get BC, and GD with trig. and all the lengths of FEC as a fixed fraction of ABC.

After that it starts to get tricky. This theorem may help:

http://www.mathisfunforum.com/viewtopic … 92#p368592

but so far that has only led me to some complicated algebra which might work if I kept going for long enough.

Oh yes, I forgot. Mark the centre and some radii.

Bob

hi lovealgebranotgeometry

Welcome to the forum.

Quick clarification needed as I don't use the 'measure of an arc' notation. When you say the measure of an arc is 40, do you mean that it is 40 units long, or that the arc makes an angle of 40 degrees at the centre of the circle. In your work above, you appear to mean the latter.

Thanks,

Bob

LATER EDIT:

Haven't worked out EF yet. I'm working on similar triangles.

hi thickhead

Agreed!

Please don't misunderstand me. I found your earlier post very helpful. I couldn't see where 3/5 came from until then. The difficulty lies with the wording of the question. If it said " A class is chosen at random, and then a student within that class is chosen at random", then I would be happy with 3/5 as an answer.

But the question only says a student is chosen at random, so that makes each of the 45 students equally likely to be picked. Knowing a girl has been picked further reduces us to 22 of the students. The number of boys becomes irrelevant.

When exam questions are made in the UK, the paper is then checked by someone else independently, in order to try to avoid this sort of thing. I wonder which book this came from?

Bob

The correct formula is

If we assume that P(A) = P(B) then this reduces to thickhead's formula and the result 3/5 follows.

But the question does not say that a class is chosen first with equal probability.

If a West High precalculus student chosen at random

The classes are unequal so P(A) is not equal to P(B).

The correct calculation is

Bob

hi markosheehan

I agree that a=0. But you'll have to consider y direction too if you want to complete this question,

I don't see where you got that value of k from. It must be wrong as k cannot be negative.

Bob

hi davidtrinh

The decibel scale is logarithmic** so to add two sounds you need to 'anti-log' the amounts then re-log the result.

https://en.wikipedia.org/wiki/Decibel

decibel is one tenth of a bel so you need to convert first:

80 db = 8 b

so compute 10*log(10^8 + 10^8)

This comes out as 83 db

** so if the db goes up by +10 the sound is 10 times as loud.

Bob

You could insert an image just above the line.

Like this (except maybe smaller ... this was just one I had handy):

Bob

hi markosheehan

You can use a similar approach to the previous one. Take x as along the line of centres. I used Wx and Wy for km before ; Ux and Uy for m before ; Vx and Vy for km after.

You can make 4 equations: conservation of momentum in each direction; experimental law in the x direction; no change of relative velocity in the y direction*.

This last will allow you to show that Uy is zero. Looks like you already have Wx is zero.

Bob

* These problems are the first time I've had to use this 'no change' law, but I think it makes sense in terms of Newton's laws. In the y direction the spheres just scrape past each other without any 'bounce' so why should their relative velocities change.

hi tonyjaa,

There are two skills at work here: (1) having a pair of brackets and multiplying them to get a sum of terms and (2) reversing this by starting with the (quadratic) expression and putting it back into brackets as a pair of factors. I think it is easier to learn to do (1) before (2)

You'll find (1) here: http://www.mathsisfun.com/algebra/polyn … lying.html

So how do we get 6x² + 19x + 15 back into brackets.

Here's a diagram for this:

6x² could have come from 1x times 6x or from 2x times 3x.

15 could have come from 1 times 15 or from 3 times 5.

Only the correct choices above will give the correct middle term of 19x. I quickly try out all the combinations to see which works.

Bob

ps. You'll notice I wrote 'times' rather than use a cross: 'x'. Unfortunately mathematicians like using 'x' for an unknown amount, and also as the multiplication sign. This could lead to confusion. So in algebraic expressions the multiply sign is often left out completely. This is why 2 times x is written as 2x. No times sign at all.

And 6 times x times x is written as 6x² (say: six ex squared)

hi tonyjaa,

I don't want to be discouraging but are you familiar with the expression "trying to run before you can walk" ?

We can get to that example eventually, but there's a lot to do first.

How much algebra have you done already? Please post an example of what you can do.

Bob

Love it!!!

And it works in IE11. Hurrah!

Bob

hi tonyjaa

If {something} times {another_something} = {yet_another_something} then {something} and {another_something} are **factors** of {yet_another_something}

eg. We know that 2 x 6 = 12. 2 and 6 are factors of 12.

In your other post about primes we looked at lots of number factors. In your algebra book you are going to look at algebraic factors

... stuff like 2y + 6z = 2 times (y + 3z)

This started with two things added together and has landed up with two things multiplied together. That is what algebraic factorisation is about. You'll notice it is in part II. I think you need to do part I first. Then it'll be easier.

Bob

EDIT AFTER I READ YOUR POST:

There's a great way to learn about solving equations here:

bobbym wrote:

Everything else you seem to have a good grasp of unless I missed something .

I think tonyjaa is quoting these definitions rather than showing they are understood.

If a number is not a prime number then it can be shown as a rectangular number**:

This diagram shows that 12 is 6x2 and 3x4. The factors of 12 are 1,2,3,4,6,12.

By splitting 12 in half you can make 2 x 6. But 6 can be further split into 2 x 3. If you keep trying to split a number like this, eventually you can split no more. At that stage you have the prime factors. Here's an example:

54 .... I know that 54 = 6 x 9. But I know I can split these numbers too. 6 = 2 x 3 and 9 = 3 x 3 ...... so 54 = 2 x 3 x 3 x 3

It doesn't matter how you do the first split; you always end up the same.

54 = 2 x 27 = 2 x 3 x 9 = 2 x 3 x 3 x 3

or

54 = 3 x 18 = 3 x 3 x 6 = 3 x 3 x 3 x 2

** 1 is a special case. If you decided to say that 1 is also a prime you'd get this

54 = 2 x 3 x 3 x 3 = (2 x 1) x (3 x 1) x (3 x 1) x (3 x 1) = (2 x 1 x 1 x 1 x 1 x 1) x (3 x 1 x 1 x 1 x 1 x 1) .... and so on.

So it's best to say 1 isn't prime and 1 isn't a rectangular number.

Every number (except 1) can be shown to be a product of some prime numbers.

example.

855 As this ends in a '5' it must have 5 as a factor so I'll start there:

855 = 5 x 171

Now, can 171 be split further? Notice that 1+7+1 = 9. There's a handy rule*** that if the digits add up to a number in the 9 times table, then the number is divisible by 9. So I'll divide 171 by 9 next.

855 = 5 x 9 x 19

Is that the end?

No, because 9 has factors.

855 = 5 x 3 x 3 x 19

Now I'm done, because all of these factors is a prime and so cannot be split any more.

I can use this to get all the factors including the non-prime ones.

5 x 3 = 15, 3 x 3 = 9, 5 x 19 = 95, 3 x 19 = 57, 5 x 3 x 3 = 45, 5 x 3 x 19 = 285, 3 x 3 x 19 = 171

So a complete list of factors is {1, 3, 5, 9, 15, 19, 45, 57, 95, 171, 285, 855}

If you can do this for any number, you'll find it very useful in all sorts of arithmetic.

*** This rule for the 9 times table also works for 3 times table, but it doesn't work for the others.

Bob

hi zapyourtumor

Welcome to the forum.

This is a long thread and there's at least one other on this question. The answer to your question is in there somewhere; you just need to go back to the early posts. I recommend you use my labelling rather than using M and N. Then you'll find the answer.

Bob

hi bossk171

And what was the final result?

I'm still thinking about thickhead's post. I feel he is correct but I don't yet understand fully why this is so.

Bob

hi Maharadja

OK. I need to put more detail to the story.

A catalogue is a special book; it consists of a list of books of a certain type. For example, if you are interested in Einstein's theory of relativity you might have a look at 'the catalogue of all books about Einstein's theory'. Saves time and allows you to make a good choice about which to read first.

But catalogues can be lists of any type of book. I'm not saying it's such a useful catalogue, but the one that lists all the red books can exist in my imaginary library if I say so. They are part of the library.

So when we look inside 'the catalogue of all books written in English', one of the entries is this catalogue itself. It is an example of a catalogue that contains itself.

'The catalogue of all the books in the library' is another self containing catalogue as it is a book and it is in the library.

'The catalogue of all catalogues that contain themselves' is another example. This catalogue actually lists itself, which means it qualifies as a catalogue that contains itself .............. but, that's ok because it does contain itself. No contradiction there.

The tricky one is 'the catalogue of all catalogues that don't list themselves'. Let's take a guess and say it doesn't have itself in the lists. In that case it satisfies the criterion for inclusion so it should contain itself. But if it does contain itself then it is one of the ones that don't contain itself. This is the paradox.

Here's another way of looking at it.

Fred is a barber and he shaves every man who doesn't shave themselves. So does he shave himself?

More importantly, how do you know the librarian is a female?

A good question. In the UK, it used to be the case that one would say 'he' in giving examples. But we have had an equality revolution and so I thought I should make an effort not to do this. So my librarian is a lady. Fred is a man, so there's a balance here. You might have spotted a flaw in the barber example. If so, keep it to yourself; it's only a story anyway.

Bob

hi Maharadja

R is a set containing only sets so you might like to start with this:

You can explain the paradox without set notation at all so maybe the following will help.

A librarian makes a series of catalogues like these:

The catalogue of all books with a red cover.

The catalogue of all books written in English.

etc etc.

She notices that some catalogues contain themselves in the listing. For example: 'the catalogue of all books written in English', is written in English so it contains itself. However, the catalogue of all books with a red cover does not have a red cover, so it does not contain itself.

She decides to make a new catalogue; the catalogue of all catalogues that do not contain themselves. 'Books with a red cover' is one entry.

So now we can consider the paradox.

Should 'the catalogue of all catalogues that do not contain themselves', contain itself? If it does then it shouldn't, and it if doesn't, then it should.

Bob

hi tonyjaa

Welcome to the forum.

There is a second website called http://www.mathsisfun.com/

It has many pages that teach mathematical topics. At the bottom of each page you will find practice questions. Also the pages are 'cross-linked' so if you come across a term you want to know more about, it is easy to find the page for that.

Hope that helps,

Bob

hi bossk171

I agree with you. I used a probability tree and arrived at the same as you did using Bayes.

The ultimate test of a probability is to run a simulation. I used BBC BASIC because it's fairly easy to follow the code.

10 DIM A$(45)

20 FOR i = 1 TO 45

30 READ A$(i)

40 NEXT i

50 countG = 0:countA = 0

60 FOR j = 1 TO 10000000

70 p = RND(45)

80 IF RIGHT$(A$(p),1) = "G" THEN countG = countG + 1 :IF LEFT$(A$(p),1) = "A" THEN countA = countA + 1

90 NEXT j

100 PRINT countA/countG

999 END

1000 DATA AG,AG,AG,AG,AG,AG,AG,AG,AG,AG,AG,AG

1010 DATA AB,AB,AB,AB,AB,AB,AB,AB

1020 DATA BG,BG,BG,BG,BG,BG,BG,BG,BG,BG

1030 DATA BB,BB,BB,BB,BB,BB,BB,BB,BB,BB,BB,BB,BB,BB,BB

This sets up the 45 data items and then runs a simulation 10000000 times. (less than a minute on my laptop)

It identifies just the girls 'found' and then whether they're in class A.

I got 0.545576.... which is a lot closer to 12/22 than to 3/5.

The only thing I can think of to explain this is that the questioner didn't mean exactly what is written in the question. Cannot think of an alternative that will give 3/5 though.

Bob

hi MathsIsFun

Works for me in Chrome but no graph area or edit button in IE11.

Bob