- Tue Oct 19, 2010 5:38 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1
- Replies:
**10** - Views:
**6471**

Yes . Sorry, but I suppose it is more complicated. It is really simple and thanks a lot. Now, go to the Geometry section, I have a new task. Please help on it!

- Tue Oct 19, 2010 4:27 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1
- Replies:
**10** - Views:
**6471**

Sorry, but it is not so helpful!

Please give me step by step procedure.

Please give me step by step procedure.

- Tue Oct 19, 2010 2:47 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1
- Replies:
**10** - Views:
**6471**

You mean that this equation x/a=y/b=z/c should be split in two equations such as x/a=y/b and y/b=z/c. So x= (az)/c and y=(bz)/c. Does it what you mean?

- Tue Oct 19, 2010 1:49 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1
- Replies:
**10** - Views:
**6471**

But what is a purpose of plugging into the equation of xy + yz + zx? Actually, how to reach this equation?

- Tue Oct 19, 2010 7:47 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1
- Replies:
**10** - Views:
**6471**

And then? How to include this in the first equation x/a=y/b=z/c ?

- Mon Oct 18, 2010 8:00 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1
- Replies:
**10** - Views:
**6471**

Help to solve this:

If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0

If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0

- Wed Oct 13, 2010 8:06 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: 11 consecutive compounded numbers
- Replies:
**9** - Views:
**2903**

Or, possible this should be solution: 12!+2, 12!+3, 12!+4, 12!+5, 12!+6, 12!+7, 12!+8, 12!+9, 12!+10, 12!+11 and 12!+12 , (total 11 consecutive numbers) since each number could be extracted by a number behind the “+” (e.g. +2, +3, +4….+12) and divided with the same number 2, 3, 4, 5,…. 12 respective...

- Tue Oct 12, 2010 2:31 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: 11 consecutive compounded numbers
- Replies:
**9** - Views:
**2903**

Yes, they are consecutive numbers, it is provided by the second part of the "compounded" number, e.g. (n+2)!+2 by ..+2, ..+3, ..+4, ..+5,.... ..+12.

- Tue Oct 12, 2010 12:55 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: 11 consecutive compounded numbers
- Replies:
**9** - Views:
**2903**

Possible I have solution, but I would like to be confirmed from others. (n+2)!+2, (n+3)!+3, (n+4)!+4, (n+5)!+5, (n+6)!+6, (n+7)!+7, (n+8)!+8, (n+9)!+9, (n+10)!+10, (n+11)!+11 and (n+12)!+12 , since each number could be extracted by a number in the brackets and divided what it should be a prove that ...

- Mon Oct 11, 2010 1:32 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: 11 consecutive compounded numbers
- Replies:
**9** - Views:
**2903**

compunde number is number which is possible to bi divided as natural number, oppoite from the simople number such as 1, 3, 5, 7, 11.