There's again one small detail on which I'm not sure. (Proposition 44 - book 1)

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI44.html

Here's the quote :

"Then HLKF is a parallelogram, HK is its diameter, and AG and ME are parallelograms, and LB and BF are the so-called complements about HK. Therefore LB equals BF."

So, we have constructed a parallelogram HLKF and we have also constructed the diameter HK which bisects it. Now, the only part which I'm not sure is concerning the small parallelograms AG ME and the complements LB BF. We know that by propositon 43, the complements of a parallelogram equal one another. My question is : knowing that I have parallelogram that has a bisector, do we automatically know that there are two parts which are smaller parallelograms and that there are two other parts which are complements ? Can we just assume that these four smaller parts exists without proving it in proposition 44 ? Is saying "ok, here are two complements and two smaller parallelograms) enough ?

Thank you!

]]>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

]]>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

Bob, thank you very much for your help! Trust me, I will be back Very soon. It's challenging for me to learn math on my own, but it's the only option I have now.

Have a grate one!

Bob

]]>B

]]>Bob

]]>Bob

]]>Welcome to the forum!

I guess you're referring to the puzzle found here:

What 5-digit number has the following features:

If we put the numeral 1 at the beginning, we get a number three times smaller than if we put the numeral 1 at the end of the number.

The answer and the solution method (algebraic) are given here.

Here's another approach, one that doesn't use an algebraic equation:

Let the 5-digit number be *abcde*, where each letter represents one of the 5 digits.

From this, the number *beginning* with 1 is 1*abcde*, and the number *ending* with 1 is *abcde*1.*Note: The fact that the number's letters are different from each other doesn't necessarily mean that the digits are all different.*

Work backwards, digit by digit, to find which digit values result in the following:

1*abcde*

x3

--------*abcde*1

=====

Start by asking this question to find *e* in 1*abcde* (and hence in *abcde*1):

The product of which digit when multiplied by 3 ends in 1?

Answer: 7 (ie, 7x3=21)...so *e*=7 (carry 2...ie, the product's tens unit)

To find *d, c, b and a*, ask the same question as for *e*, but this time there's a 'carry' to add. eg, to find *d*:

The product of which digit when multiplied by 3, plus the carry 2, ends in 7?

Answer: 5 (ie, (5x3)+2=17)...so *d*=5 (carry 1)

So far we have this partial solution:

1*abc*57

x3

--------*abc*571

=====

...and the rest I'll leave for you to solve.

]]>Bob

]]>I am getting

can you solve for h1 and h2 now?

]]>