Question

HELP ME PLEASE!!!
Divide3K^2/4K-2 / K^2+3K+2/2K^2+5K-3

Answer 1

@timo86m help?

Answer 2

is your question
divide \[\frac{ 3k ^{2} }{4k-2 } by \frac{ k ^{2}+3k+2 }{2k ^{2}+5k-3 } ~?\]

Answer 3

yes please

Answer 4

I think that is what she meant surji.

Use perenthesis next time pinky.

and first step is to take the reciprocal of the second ratio :D

Answer 5

http://openstudy.com/users/timo86m#/updates/52fe7476e4b0332f69f7dec3

THis is similar pink mommy

Answer 6

that whole thing confuses me.

Answer 7

Can you show me how to do this one step by step please?

Answer 8

well first of all use perenthesis and spaces makes it easier to read

((3K^2)/(4K-2)) / ((K^2+3K+2)/(2K^2+5K-3))


here is a template

( ( )/( ) ) / ( ( )/( ) )

Answer 9

when in doubt use perenthesis for groupings

Answer 10

@timo86m sorry I didn't know how to write it correctly.

Answer 11

\[\frac{ 3k ^{2} }{ 2\left( 2k-1 \right)}\times \frac{ 2k ^{2}+5k-3 }{ k ^{2}+3k+2 }\]
make factors and then cancel if you have common factors.

Answer 12

It's ok but it is very important :) cuzz if you know how then you can use wolfram. It jsut makes it able for computers and humans to read it :)

Answer 13

((3K^2)/(4K-2)) / ((K^2+3K+2)/(2K^2+5K-3))

first thing is to take reciprocal of second ratio: I do it below and notice the / is now a * :)

((3K^2)/(4K-2)) * ((2K^2+5K-3)/(K^2+3K+2))




i will try is to factor them into binomials leaving noe k^2 but only ks: as shown below


((3K^2)/(4K-2)) * ((2K^2+5K-3)/(K^2+3K+2))
^ ^ already factored ^ ^


try it out :)

Answer 14

if yo ucant factor it leave it alone and move on to next one. Like 3k^2 i think you cant so move on to the next one :)

Answer 15

(k+3)(2k−1) /(K+2)(K+1)

Answer 16

2*K^2+5*K-3 = (K + 3) (2 K - 1)


K^2+3*K+2 = (K + 2) (K + 1)


try to them them cancel :)

Answer 17

nothing to cancel