Algebra help.

[Simpleboy]

So, Im stuck. Help me with these questions.

Prove that the arithmetic mean of two numbers is not less than their geometric mean. hence, show that if teh sum of four positive numbers is 1, their product is not more than 4^-4

If a:b = c: d prove that a+c:a+b+c+d = a:a+b

if a and g are the arithmetic and geometric means respectively of 'n' and 'm, and k^2 is the arithemetic mean of m^2 and n^2, prove that a^2 is the arithmetic mean of g^2 and k^2

Any helps appreciated.

[Crism]

47!

[indian airlines]

Ok, this I think is the answer for the one you asked me on MSN. The question asked for values of x which arose as a result of different values of k, so here are the 4 values of k and their corresponding x values.

1. If k=0, then -inf<x<inf 

2. If k=x, then -inf<x<inf 

3. If k>0, then:

            a) if x>k, then x>0

            b) if x<k, then x has no real values 

4. If k<0, then:

            a) if x>k, then -inf<x<inf

            b) if x<k, then x has no real values 

I had to put this in the code boxes because for some reason the VB code was deleting certain things I had written.

Ok, this is the answer for the second one you posted up there:

 a/b = c/d. Cross multiplying, we get: ad=bc. Let this be equality #1.
 
You need to prove that (a+c)/(a+b+c+d) = a/(a+b)
 
So now, cross multiplying this equality, we get:
 
aa+ab+ac+bc=aa+ab+ac+ad
 
Cancelling like terms from both sides, we are left with:
 
bc = ad
 
But bc and ad are equal as shown in equality #1 earlier, therefore:
 
(a+c)/(a+b+c+d) = a/(a+c) 

Hope that wasn't too complicated.

[Mungous]

Sounds to me like your being taught something you don't need to be

[Simpleboy]

Yeah, the accelerated class sucks at times.

[Cam]

Yeah, the accelerated class sucks at times.

I was going to post, but I was in the Decelerated class.