Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Simplify completely (x^2+x-12)/(x^2-x-20)÷(3x^2-24+45)/(12x^2-48x-60)

OpenStudy (anonymous):

i got 4(x+1)/(x-5)?

OpenStudy (anonymous):

help?

sam (.sam.):

(x^(2)+x-12)/(x^(2)-x-20){(3x^(2)-24+45)/(12x^(2)-48x-60) ((x+4)(x-3))/(x^(2)-x-20)*(12x^(2)-48x-60)/(3x^(2)-24+45) ((x+4)(x-3))/((x+4)(x-5))*(12x^(2)-48x-60)/(3x^(2)-24+45) (x-3)/(x-5)*(12x^(2)-48x-60)/(3x^(2)-24+45) (x-3)/(x-5) * (12x^(2)-48x-60)/(3x^(2)+21) (x-3)/(x-5) * (12(x^(2)-4x-5))/(3x^(2)+21) (x-3)/(x-5) * (12(x+1)(x-5))/(3x^(2)+21) (x-3)/(x-5) * 12(x+1)(x-5))/(3(x^(2)+7)) (x-3)/(x-5) * 4(x+1)(x-5)/(x^(2)+7) (x-3) * 4(x+1)/(x^(2)+7) 4(x-3)(x+1)/(x^(2)+7)

OpenStudy (anonymous):

its multiple choice A- (x-5)/4(x+1) B- (x+1)/8(x-5) C-4(x+1)/(x-5) D-8(x-5)/(x+1)

OpenStudy (anonymous):

I'm sorry. It's You're correct. It's C

OpenStudy (anonymous):

(: what about this one

OpenStudy (anonymous):

(x^2-4x+4)/(x^2+10x+25) * (x+5)/(x^2+3x-10)

sam (.sam.):

(x-2)/(x+5)^2

OpenStudy (anonymous):

x-2/(x+5)^2

OpenStudy (anonymous):

(x^2-100)/(x^2+2x-120) ÷ (6x+60)/(x-2)?

OpenStudy (anonymous):

o.0 i dont understand how you can do it this fast

OpenStudy (anonymous):

sorry, x+10/x+12

OpenStudy (anonymous):

x^2-100 = (x+10)(x-10) x^2+2x-120= (x+12)(x-10) ~~ I have a trick to get this fast

OpenStudy (anonymous):

Multiple choice again x-1/6(x+6) x-2/6(x+10) x-2/6(x+12) x-1/3(x+12)

OpenStudy (anonymous):

Oh, Ok. Sorry. I only did the first part.

OpenStudy (anonymous):

x-2/6(x+12)

OpenStudy (anonymous):

one more question (x^2+4x-5)/5x^2-8x+3) * (20x-12)/x^2-6x-55) A) 4/x+5 B)4(x-1)/x+5 C) 4/x-11 D)4(x+5)/x-11

OpenStudy (anonymous):

C) 4/x-11

OpenStudy (anonymous):

Hope you got it :)

OpenStudy (anonymous):

Thank you so much for all your help I'm sure ill be back haha

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!