x^2 - x - 20/x^2 - 3x - 10 * x^2 + 7x + 10/x^2 + 4x - 5
2007-02-28 02:24:09 · 7 answers · asked by spanishfly_100 1 in Science & Mathematics ➔ Mathematics
[(x^2-x-20)/(x^2-3x-10)]*[(x^2+7x+10)/(x^2+4x-5)]
x^2-x-20=x^2-5x+4x-20=(x+4)(x-5)
x^2-3x-10=x^2-5x+2x-10=(x-5)(x+2)
x^2+7x+10=x^2+5x+2x+10=(x+5)(x+2)
x^2+4x-5=x^2+5x-x-5=(x+5)(x-1)
=[(x+4)(x-5)/(x-5)(x+2)]*[(x+5)(x+2)/(x+5)(x-1)]
=(x+4)/(x-1)
2007-02-28 02:34:04 · answer #1 · answered by Maths Rocks 4 · 0⤊ 0⤋
x^2 -x -(20/x^2) -3x -10x^2 +7x +(10/x^2) +4x-5
=-9x^2 +7x -(10/x^2) -5
= -9x^4 +7x^3 -10 -5x^2
= -9x^4 +7x^3 - 5x^2 -10
2007-02-28 10:44:34 · answer #2 · answered by jaybee 4 · 0⤊ 0⤋
Impossible
2007-02-28 10:26:43 · answer #3 · answered by Anonymous · 0⤊ 0⤋
I have no idea, but you could get tutoring and ask a teacher for help. Getting the answer from someone on here and acting like you know it is one thing, but asking the teacher and showing them you really care may help bring your grade up anyway. Sorry, didn't mean to preach, have a good day, and try to do your best.
2007-02-28 10:28:09 · answer #4 · answered by Anonymous · 0⤊ 0⤋
-19x^2 +7x - 5
2007-02-28 10:27:09 · answer #5 · answered by Anonymous · 0⤊ 0⤋
2^ cannot be preceeded by a non-intrisic value (example 20^ * -5), re-write the problem and re-submit it.
2007-02-28 10:27:49 · answer #6 · answered by john y 3 · 0⤊ 0⤋
dam
2007-02-28 10:26:32 · answer #7 · answered by Anonymous · 0⤊ 0⤋