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Mathematics 21 Online
OpenStudy (anonymous):

How do you figure out this for factor by grouping 6x^3-6x^2-x+1

OpenStudy (anonymous):

(x-1) (6x^2 -1)

OpenStudy (anonymous):

if u like the answer ...pls fan me ..thanx

OpenStudy (anonymous):

then I have another one to see if it is correct I have 6^6-64

OpenStudy (anonymous):

my answer (b+2)(b^2-2b+4)(b-2)(b^2-2b+4)

OpenStudy (anonymous):

r u typing?

OpenStudy (anonymous):

hello, r u there?

OpenStudy (anonymous):

yes I am here

OpenStudy (anonymous):

Small ever in that factorisation: one of the (-2b) should be +2b

OpenStudy (anonymous):

i m waiting for the question

OpenStudy (anonymous):

error* eugh

OpenStudy (anonymous):

b^6-64

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

OK, Ignore me then... ¬_¬

OpenStudy (anonymous):

its (b-8)(b+8)

OpenStudy (anonymous):

because of (a^2 - b^2) =( a-b) (a+b)

OpenStudy (anonymous):

it should (b+2)(b^2-2b+4)(b-2)(b^2+2b+4)

OpenStudy (anonymous):

hey !!! was that squared or raised to power of 6 ? 6???

OpenStudy (anonymous):

Yes, that is right.

OpenStudy (anonymous):

ok i misread it

OpenStudy (anonymous):

sry

OpenStudy (anonymous):

no problem

OpenStudy (anonymous):

thank you INewton

OpenStudy (anonymous):

No problem.

OpenStudy (anonymous):

tht math looks confusing but im just watching cuz im going 2 hav 2 do this sometime in life

OpenStudy (anonymous):

u will be fine macy....its not really hard...

OpenStudy (anonymous):

ok thanks ;)

OpenStudy (anonymous):

i cant type in group chat :(

OpenStudy (anonymous):

oh y?

OpenStudy (anonymous):

Reload the website.

OpenStudy (anonymous):

smtimes ...it just doesnt work...

OpenStudy (anonymous):

ohhhh. or u can refresh it wen tht hapens i refresh and it works

OpenStudy (anonymous):

didnt work for me.....:((

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

macy r u on gmail or yahoo?

OpenStudy (anonymous):

im nt aloud 2 give my email on here either

OpenStudy (anonymous):

ok because its not really good to talk in someone's question...thats all..

OpenStudy (anonymous):

hmmmmm im thinken and thinken and drawing a blank

OpenStudy (anonymous):

wait IDEA!!!!

OpenStudy (anonymous):

hahah

OpenStudy (anonymous):

wat?

OpenStudy (anonymous):

either me or u can make a question like hey or something and talk on tht

OpenStudy (anonymous):

i feel smart

OpenStudy (anonymous):

hahah......u r really smart

OpenStudy (anonymous):

awww thanks

OpenStudy (anonymous):

go 2 the next question thts posted ok?

OpenStudy (anonymous):

u there?

OpenStudy (anonymous):

I have a question polynomial 0.041h-0.018a-2.69 can be used to estimate the lung capacity, in liters of a female with height h, in centimeters, and age a, in years. Find the lung capacity of a 30 year old woman who is 170 centimeters tall. Her lung capacity is approximately_____ liters?

OpenStudy (anonymous):

I am confused on how you go about figuring out this problem

OpenStudy (anonymous):

Lung Capacity (h, a) = 0.041h-0.018a-2.69 Let h = 170 , a = 30 => Lung Capacity (170, 30) = 0.041 x 170 - 0.018 x 30 - 2.69

OpenStudy (anonymous):

the answer would be 206 liters

OpenStudy (anonymous):

0.041 x 170 - 0.018 x 30 - 2.69 = 3.74...

OpenStudy (anonymous):

ok i see what i did wrong

OpenStudy (anonymous):

Was your mistake considering the possibility that my advice was incorrect?

OpenStudy (anonymous):

what i did was took 0.041* 170=6.97-.018=6.952*30=208.56-2.69=205.87

OpenStudy (anonymous):

Ah, I see; glad you sorted it.

OpenStudy (anonymous):

but if you do the mutliplication first 0.041*170=6.97 & .018*30=.54 then you subtract 6.97-.54-2.69=3.74

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