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I don't know my Euclidean propositions as well as you, but I suggest you look back to prop. 34.

There are a number of ways of defining a parallelogram:

All definitions require 4 sides but then:

opposite sides equal in length

OR

opposite sides parallel

OR

one pair of opposite sides equal and parallel

OR EVEN

diagonals bisect each other

I expect there are more. You can quickly show these are equivalent by congruent triangles.

Bob

That MOD in histogram should be MODE. It has no connection with MOD (apart from sharing some letters)

Bob

From what I understand, there is little difference between the two methods. Both rely on alternate angles. In Euclidean geometry there is often more than one way to prove something. As long as he only uses axioms and theorems that have already been proved, what does it matter ?

Bob

Yes. If you draw one diagonal to divide the quadrilateral into two triangles, it is not too hard to show these are congruent (side and two angles). Hence opposite sides are equal in length.

Bob

I've had a quick look at your other post but it will take a while to take in all the relevant propositions. I'll reply when I've had time to do this.

Yes and replied.

B

From the graphs (using trial to 'home in' on the value of a) it looked like this single point of contact was at (e , e) so I used differentiation to get an expression for the gradient and set this equal to 1. Then using (e , e) I was able to get 'a'. Substituting back shows this value is right which is what ElainaVW meant by 'ansatz'. I cannot find an algebraic way to derive a value for a.

Bob

hi ducmod

I'll be happy to provide further help. You'll need to understand the formula for variance and co-variance. The book I have used to teach from is called Advanced Level Statistics by A Francis. ISBN 0-85950-451-4 My book was printed in 1986 so it may not be available now.

Post further questions when you are ready.

Bob

hi ElainaVW

Thanks. I just logged in to say that as well. Except I don't know why. I've just got it by trial.

Bob

hi Nemexia

Thanks for clearing that up. I'm getting a value of about a = 1.445 but only by looking at the graphs. I will try to find a better method. Here is what I did:

and

Then I used http://www.mathsisfun.com/data/function-grapher-old.php

to get these:

Here you can see the graphs when a = 1.2 and when a = 1.5

I tried different a values until the graphs just touched. I'll keep thinking and try to come up with a way to calculate this a exactly.

LATER EDIT:

These functions are inverses of each other so we want the graph where the line y=x makes a tangent to the curve. ie gradient angle = pi/4.

Got to go out now, so I'll finish this off when I get back.

Bob

hi Al-Allo,

He says: "If AB does not equal DE, then one of them is greater. Let AB be greater."

So he chooses to letter the vertices so that this is true. Therefore he doesn't need to consider the other case.

It is like this:

Two people decide to share their money equally.

If they have the same already, do nothing as they have achieved their aim.

If not, let A be the one with more and B the other. Let A's money be a and B's money be b.

Calculate (a-b)/2. A gives B this amount.

Now A has a - (a-b)/2 = (2a - a + b)/2 = (a + b)/2 and B has b + (a-b)/2 = (2b + a - b)/2 = (a + b)/2 so they have the same.

It doesn't matter what their real names are; we choose A to be the label for the one who has more.

Bob

hi EbenezerSon

A cyclist starts a journey from town A. He rides 10km north, then 5km east and finally 10km on a bearing of 045.

(a) How far east is the cyclist's destination from town A?

(b) How far north is the cyclist's destination from A?

(C) Find the distance and bearing of the cyclist's destination from town A.

(Correct your answers to nearest kilometers and degree)

Total Easterly component = 0 + 5 + 10sin(45)

Total Northerly component = 10 + 0 + 10cos(45)

This should work out OK. Post answers please.

Bob

Un peu. Mais je suis un Francophile.

Bob

je vous en prie

Bob

As far as I remember, Bob Bundy made it clear to me that not all bearing questions could be solved using vector approach

Did I say that ? I've searched old posts and found only one on this topic. I did say then that your calculator would not sort out the correct angle without you referring to a diagram, but the approach still used vectors. I look forward to the question.

Bob

hi Nemexia

What do you mean by f(x)=log(a,x)

Is this

?

Bob

hi math9maniac

math9maniac wrote:

1) Two circles with centres A and B, intersect at X and Y. Show that the common chord XY is the perpendicular bisector of the line AB joining the centres.

Did you mean AB is the perpendicular bisector of XY ?

In any circle, centre A, draw two radii, AX and AY. As AX = AY, the triangle AXY is isosceles. Let M be the midpoint of XY. What can you say about triangles AMX and AMY ? And hence about the angles AMX and AMY ? ........................

Bob

Have you looked back to prop 16 ? Why cannot you just turn the triangle to bring a fresh angle into the required position, re-lettering the points as required?

Bob

hi ducmod

Welcome to the forum.

It sounds like you are describing the Pearson correlation coefficient. (There are others.)

There's a wiki page but, if you are new to this subject, you may find it difficult.

http://en.wikipedia.org/wiki/Pearson_pr … oefficient

There's a lot of theory behind that formula. I'll try to give you an outline.

If you take your x and y pairs and plot them on a graph you'll get a scatter graph. You can get a good idea of how correlated two variables are just by looking at the points. If they are randomly scattered then there's no correlation at all. If they lie in a straight line, there's a perfect correlation between the two and you could work out the equation of the line in the form y = mx + c.

That's unlikely in practice; the points may show a tendency towards a straight line but with not all the points lying exactly on a line. The Pearson correlation coefficient is a way to calculate a number (always between -1 and + 1) that indicates how close to the line the points are.

Pretend you had the line drawn on the scatter graph. For each point calculate how far above or below the line it lies. Square this distance to ensure all these measures are positive. Now, add them up. The resulting sum gives a measure of how well that line fits the points. If you change the line, all the calculations lead to a new sum. The Pearson CC gives the line when the sum of squares is a minimum. In summary it's the line when the distances (squared) are the lowest they could be.

So there's a load of algebra for calculating these distances, then squaring and adding. What the minimum gives is the 'm' and 'c' for the best fit line*. From that you can generate a number that has the desired property (-1 < p < +1) and that's why the formula has that form.

It's a part of regression analysis; or method of least squares. I cannot find either on the Maths Is Fun site but there are bound to be loads of on-line sites that cover this. It's a while since I taught this, so I'd have to remind myself how it all goes. But If you're keen, I'll have a go. I'll warn you now, it will take a few posts to cover it all and the algebra is tough (well it is for me anyway ). Post again if you want to know more.

Bob

* It's called the Y on X regression line and can be used to predict a fresh Y given an X. Strictly, if you wanted to predict an X from a Y you should calculate the X on Y line, which is done by calculating horizontal distances from the points to the line.

Ln(4) _ ln(2) + ln(3) = ln(4 x 3/2) = ln(12/2)=ln(6)

Add the plus and divide the minus

Bob

If you take a cylinder and "unroll" it you get a rectangle. The width is equal to the circumference of the top ie. 2 x pi x radius and the height is the height of the cylinder

Bob

Thanks for clearing that up.

If A = [w,x,y,z] where w,x,y,z ∈ {0,1} then there are 16 possible elements in the space for A ( [0,0,0,0] through to [1,1,1,1]

Similarly for S.

The choice is RANDOM, so the probability of any particular event is 1/16 in each register and 1/16 x 1/16 = 1/256 for each output.

Some outputs may occur from more than one starting point and maybe that is where this question is going.

The definition of independent events is P(A and B) = P(A).P(B) ( sorry, I cannot find a quick way to make an 'intersection' symbol )

So what does P(A and B) means exactly?

It should mean the probability that A and B both occur in which case:

eg. A = [0,1,0,1] and S = [0,1,0,1] gives output = [1,1] and so does A = [1,1,1,1] and S = [0,1,0,1] but that does not effect the independence of the inputs; it just makes output = [1,1] more likely than eg. output = [1,1,1,1] which can only happen in one way.

But if P(A and B) means the probability of a particular output, then the above shows that P(A and B) ≠ P(A).P(B)

So, in your position, I'd go back to my teacher and ask what he means by P(A and B).

Bob

hi subait

Welcome to the forum.

I do not understand this at all.

subait wrote:

let assume that we have two register A and S with 0s and 1s only. Both of the two register give output at the same time (ai,si) . The output is ai if the output of the si is ‘1’ else the output of ai is discarded. Is the A and S register are independence and prove it is they are independence.

You say the "two registers give output at the same time", but then say "The output is ai ...."

It makes more sense if it reads "Both of the two register have input at the same time (ai,si)

Then (with my choice of entries) we have this:

Next, what does "discarded" mean. You cannot have empty binary digits. They have to be 1s or 0s. Did you mean 0s for these outputs?

In that case OUTPUT = A **AND** S.

And what is independence in this context? I have never met this word used in computer science. I have searched the internet and can find no reference to it. It is implicit in logic theory that the inputs are regarded as independent. It is assumed rather than proved.

So I cannot understand what this question is about at all. I am so sorry . Would you be able to post the problem exactly as worded when you met it?

Bob

hi Agnishom,

If you write

a + (p-1)d = r[a + (q-1)d] and two more similar, and then subtract you get

d[(p-1) - (q-1)] = rd[(q-1) - (r-1)] and so on.

All you need to do is simplify this.

Bob

hi sandeep8557981588

Welcome to the forum.

Bob

hi bobbym,

If p = -0.24 and x = -0.064

x + 3p + 1 = - 0.064 - 0.72 + 1 = 0.216 so cube root = 0.6

and cube root of x = - 0.4 so LH expression = 0.6 -- 0.4 = 1

Bob