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Mathematics 19 Online
OpenStudy (anonymous):

( (1/7)x^2 - (2/5)) * ( (1/2)x + (1/2) )

OpenStudy (apoorvk):

multiply each term of the anyone binomial into the other. Like this: (a+b)(c+d) = a(c+d) + b(c+d)

OpenStudy (anonymous):

so..[ (1/7)x^2 * ((1/2)x^2) + (1/2)) ] + [ ((2/5) * ((1/2)x^2) + (1/2)) ] ?

OpenStudy (anonymous):

yup thts right..

OpenStudy (apoorvk):

good except that the (2/5) has a 'minus' sign with it.

OpenStudy (anonymous):

\[\left(\begin{matrix}1 \\ 14\end{matrix}\right)x ^{3}+\left(\begin{matrix}1 \\ 14\end{matrix}\right)x ^{2}-\left(\begin{matrix}2 \\ 10\end{matrix}\right)x-\left(\begin{matrix}2 \\ 10\end{matrix}\right)\]

OpenStudy (apoorvk):

@math456 oop - that notation is actually used for 'combination' or 'C' :P. Other wise it's good.

OpenStudy (anonymous):

crap im so lost.

OpenStudy (anonymous):

ohh yea i know i couldnt find the division sign thts y i use this..

OpenStudy (apoorvk):

forget that comment, @musicalrose , you're almost there, just keep in mind that it's a "-2/5" over there, so you include the 'minus' sign as well while multiplying.

OpenStudy (anonymous):

@musicalrose i will make it clear.. u use the same concept..

OpenStudy (anonymous):

[ (1/7)x^2 *((1/2)x+(1/2) ] - (2/5*((1/2)x^2) + (1/2)) ] ?

OpenStudy (anonymous):

yes itd be written like that

OpenStudy (anonymous):

.07x^3 + 1/2 for the first part?

OpenStudy (anonymous):

(1/14)x3+(1/14)x2−(2/10)x−(2/10) <-- like this

OpenStudy (anonymous):

oooh because you made like denominators?

OpenStudy (anonymous):

yup... u'll get 0.07 for 1/14

OpenStudy (anonymous):

so shouldnt the 1/14x2 actually be 7/14 since it was 1/2x2? or am I not getting it lol

OpenStudy (anonymous):

|dw:1339011637725:dw|

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