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10 answers

we simply have 4 function that we have to simply .
if
f(x1) = (x^2 + 8x + 12)
f(x2) = (x^2 + 4x - 12)
f(x3) = (x^2 + 6x + 8)
f(x4) = (x^2 + 3x - 10)


▪f(x1) = (x^2 + 8x + 12) = ( x+6)(x+2)
▪f(x2) = (x^2 + 4x - 12) = ( x+6)(x -2)
▪f(x3) = (x^2 + 6x + 8) = ( x+4)(x +2)
▪f(x4) = (x^2 + 3x - 10) = ( x+5)(x -2)

Now put them bacl to the main formula;
if f(x1) / f(x2) / f(x3) / f(x4) so
[( x+6)(x+2) / ( x+6)(x -2)] / [( x+4)(x +2) /( x+5)(x -2) ]
in the frst part we can remove ' x+6 '
so [ (x+2) / (x-2)] / [( x+4)(x +2) / ( x+5)(x -2) ]

so here far fraction * far fraction & near fraction * near fraction

it means if
f(x1) = (x+2)
f(x2) = (x-2)
f(x3) = ( x+4)(x +2)
f(x4) =( x+5)(x -2)
so we have ; [f(x1) * f(x4)] / [ f(x2) * f(x3) ]
[ (x+2) * ( x+5)(x -2) ] / [ (x-2) * ( x+4)(x +2) ]
wow now we can remove ' x-2 ' & ' x+2 '
so we have ; x+5 / x+4 = 1( x+4 ) +1

Its our answer.
very good question , good luck.

2006-09-01 10:41:22 · answer #1 · answered by sweetie 5 · 1 1

4+4=2

2006-09-01 10:16:04 · answer #2 · answered by NONAME 2 · 1 1

(x+5)/(x+4)

The thing you need to do here is simplify the equations. Factor each of them:

(x^2 + 8x + 12) = (x+6)(x+2)
(x^2 + 4x - 12 ) = (x+6)(x-2)
(x^2 + 6x + 8 ) = (x+4)(x+2)
(x^2 + 3x - 10 ) = (x+5)(x-2)

It works out pretty easy from there.

2006-09-01 10:36:09 · answer #3 · answered by Will 6 · 0 0

(x+5)/(x+4)

2006-09-01 10:20:31 · answer #4 · answered by bretttwarwick 3 · 1 0

Are you serious this queston is so easy. Are you a second grader.

2006-09-01 10:20:04 · answer #5 · answered by Cameron C 1 · 0 0

stop posting all your homework in here boy.

2006-09-01 10:18:41 · answer #6 · answered by Anonymous · 0 1

window

2006-09-01 10:22:52 · answer #7 · answered by Anonymous · 0 1

=you find the answer yourself

2006-09-01 10:15:44 · answer #8 · answered by Anonymous · 1 1

520875632.3658

2006-09-01 10:19:37 · answer #9 · answered by Southern Apostolic 6 · 0 1

how would i know that

2006-09-01 10:18:16 · answer #10 · answered by So Wrong, It's Jessica™ 4 · 0 1

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