Search found 32 matches

Go to advanced search

by japiga
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: 5465

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Yes :clap: . 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!
by japiga
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: 5465

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Sorry, but it is not so helpful! :confused:
Please give me step by step procedure.
by japiga
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: 5465

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

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?
by japiga
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: 5465

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

But what is a purpose of plugging into the equation of xy + yz + zx? Actually, how to reach this equation?
by japiga
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: 5465

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

:confused: And then? How to include this in the first equation x/a=y/b=z/c ?
by japiga
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: 5465

x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

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
by japiga
Wed Oct 13, 2010 8:06 am
Forum: Advanced Algebra ("pre-calculus")
Topic: 11 consecutive compounded numbers
Replies: 9
Views: 2580

Re: 11 consecutive compounded numbers

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...
by japiga
Tue Oct 12, 2010 2:31 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: 11 consecutive compounded numbers
Replies: 9
Views: 2580

Re: 11 consecutive compounded numbers

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.
by japiga
Tue Oct 12, 2010 12:55 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: 11 consecutive compounded numbers
Replies: 9
Views: 2580

Re: 11 consecutive compounded numbers

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 ...
by japiga
Mon Oct 11, 2010 1:32 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: 11 consecutive compounded numbers
Replies: 9
Views: 2580

Re: 11 consecutive compounded numbers

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.

Go to advanced search