if b+c/a = c+a/b = a+b/c and a+b+c not equal to 0 then show that each of these ratio is equal to 2
A
It Could Be Anything!
You use a person's name to spell it.
One way is to show that the cross-products are equal. If A, B, C and D are four integers and B, D are not zero, then A/B = C/D if AD = BC
i will show you how to play ode to joy its like this: b b c d d c b a g g a b b a b b c d d c b a g g a b a g a a b g a b c b g a b c b a g a d b b c d d c b a g g a b a g in the third part the (b c b) are voiced together means play them fast together but the rest is slow
To show this proof, you must use substitution. If A=B and B=C, you can substitute C wherever you see B. By doing this, you can write the first equation as A=C.
Suggested layouts . . . Just play ! ( Not sure if images will show . . . If not, here they are written out . . . Layout 01 - A, B, C, D, C, B, A D, C, A, B, A, C, D A, B, C, D, C, B, A D, C, A, B, A, C, D A, B, C, D, C, B, A D, C, A, B, A, C, D A, B, C, D, C, B, A D, C, A, B, A, C, D A, B, C, D, C, B, A Layout 02 - C, D, B, A, B, D, C D, B, A, D, A, B, D B, A, D, C, D, A, B A, D, C, B, C, D, A D, C, B, A, B, C, D A, D, C, B, C, D, A B, A, D, C, D, A, B D, B, A, D, A, B, D C, D, B, A, B, D, C Layout 03 - D, B, C, B, C, B, D A, D, B, C, B, D, A D, A, D, B, D, A, D C, D, A, D, A, D, C B, C, D, A, D, C, B C, D, A, D, A, D, C D, A, D, B, D, A, D A, D, B, C, B, D, A D, B, C, B, C, B, D Layout 04 - A, B, C, D, C, B, A B, A, B, C, B, A, B D, B, A, B, A, B, D C, D, B, A, B, D, C A, C, D, B, D, C, A C, D, B, A, B, D, C D, B, A, B, A, B, D B, A, B, C, B, A, B A, B, C, D, C, B, A
A b c
(b b b)( b b b )(b d g a)(b....)(c c c c)(c b b b)(a a a b)(a...d)(b b b)(b b b)(b d g a)(b....)(c c c c)(c b b b)(d d c a)(g.....)
a b c a c a d a c c c a c b b b a a c b b b a c c b
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.