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Mathematics 11 Online
OpenStudy (anonymous):

what is equivalent to -3.5(2-3n)-2.5 -7-8n -7+8n -7-13n -7-n(10.5-2.5) -7+n(10.5-2.5)

OpenStudy (anonymous):

-2.5n

pooja195 (pooja195):

\[\huge~\rm~\bf -3.5(2-3n)-2.5n\] Ok based on what we did previously can you tell me what we would do first?

OpenStudy (anonymous):

what is equivalent to -3.5(2-3n)-2.5n

pooja195 (pooja195):

Got it :P

OpenStudy (anonymous):

lol

pooja195 (pooja195):

\[\huge~\rm~\bf~ −3.5(2−3n)−2.5n \] what's first? :#

OpenStudy (anonymous):

add 2 to each side to cancel the 2?

pooja195 (pooja195):

Try distributing to get the hard stuff outta the way :P

pooja195 (pooja195):

|dw:1451016961432:dw|

pooja195 (pooja195):

-3.5 * 2=?

OpenStudy (dayakar):

use distributive property a(b+c) = a*b+a*c

OpenStudy (anonymous):

ooh thanks bro

pooja195 (pooja195):

\[\huge~\rm~(−3.5)(2)+(−3.5)(−3n) \]

OpenStudy (anonymous):

-3/5*2 is -6 and -3.5*3n equals 9n?

pooja195 (pooja195):

its not a fraction -3.5*2=?

OpenStudy (crabbyoldgamer):

-3.5*2 is not 6

OpenStudy (anonymous):

-9n

OpenStudy (crabbyoldgamer):

-3.5 * 3n is not 9n. You multiply the number times the number.

pooja195 (pooja195):

Let's try this.. \[\huge~\rm~ -3.5 \times 2=? \]

OpenStudy (anonymous):

oh so it's -7 and -10.5?

OpenStudy (crabbyoldgamer):

yes

OpenStudy (dayakar):

2 times of - 3.5 =

pooja195 (pooja195):

Perfect now combine like terms (10.5n+−2.5n)+(−7) 10.5-2.5=?

OpenStudy (anonymous):

8

pooja195 (pooja195):

Good so we have 8n -7 remains the same What do you think our answer would be?

OpenStudy (anonymous):

n

pooja195 (pooja195):

I meant from your answer choices >_> :P

pooja195 (pooja195):

we have 8n & -7

OpenStudy (anonymous):

oh lol

OpenStudy (anonymous):

-7+8n

pooja195 (pooja195):

Yes! Well done ^_^

OpenStudy (anonymous):

thank you now should I open another one for my 4th question?

pooja195 (pooja195):

is it the last one?

pooja195 (pooja195):

If so just post it here if not make a new post

OpenStudy (anonymous):

yep

pooja195 (pooja195):

ok post it here :)

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