Use the distributive property to remove the parentheses. (y+9)11
The distributive property says \[a(b+c) = a*b + a*c\]and because of the commutative property, that means \[(b+c)a = a*b + a*c\]as well
okay so.........
Distribute the term 11 by multiplying to both terms 11(y+9) 11y+99
@theonlycool13
http://www.wolframalpha.com/input/?i=distribute+11%28y%2B9%29 Here's the proof if you don't believe me, also use this website to double-check :)
Any q's?
it was wrong thanks any way
My answer?? What was the correct one? Last time I checked the distributive property, is pretty straightforward......
okay sorry but it was wrong
No need to be sorry! I'd like the name of that teacher the distributive prop, in this case is simple multiplication! Like @whpalmer4 posted!!! thats a mathematical rule :P You should definitely protest answer
okay
so what was the right answer....
cant it is a virtual school it online
11y+99
That's the one i posted above.... here all copy/ paste no changes..... Distribute the term 11 by multiplying to both terms 11(y+9) 11y+99
The link i provided arrived to the same conclusion i did, if u click link :)
yes, there's no question that \[(y+9)11 = 11y + 99 = 99 + 11y\]
Why are you attending a virtual school?
so sorry i thought it was 11(y+9)
how was my answer wrong.... grr frustrating
ohh sorry should have made more clear... I thought you would follow my steps
no ur anser was right
oh i am so stupid
so sorry
no, no not stupid! I should have confirmed you understood, half my bad :)
this stuff is hard for me i need lots of help can you?
no yo cool
what do u need help w/?
the same kind of problem 5 in fact
go ahead and post, or u can close this q and post another and post my @Nanalew, so i can just link to it either way :)
K will :-)
10(v-3) @Nanalew
10v-30 :) You can double-check w/ the website i used earlier :)
Understand :) All u r doing is multiplying 10 to both terms :) Or in other terms distributing
i hav to eat
okay back
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