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skullpatrol Group TitleBest ResponseYou've already chosen the best response.0
What don't you understand?
 one year ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.0
Sure! basically 1(2+2) add 1 to each of the factors so you have (1+2) + (1+2) You distribute the outsider to each of the insiders. and it's not lonely anymore c: xD haha
 one year ago

skullpatrol Group TitleBest ResponseYou've already chosen the best response.0
That^ is NOT right at all
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
you multiply, not add, but the same idea yes.
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
so if the equation is x(a+b) then what you'd end up with, after distribution is ax+bx
 one year ago

skullpatrol Group TitleBest ResponseYou've already chosen the best response.0
@Profunds77 $$4( 90 + 12) = 4*90 + 4*12$$ Note 4 distributes as a MULTIPLIER of each term of the sum 90 + 12.
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
what he said
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
I like using letters instead :)
 one year ago

Profunds77 Group TitleBest ResponseYou've already chosen the best response.0
I was looking at a question answered and they mentioned the distributive rule instead of the foil method just looking for new technics that I can apply
 one year ago

Chelsea04 Group TitleBest ResponseYou've already chosen the best response.0
FOIL as in: First Outside Inside Last Are you taliking about that foil method?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
FOIL is just a memory crutch to keep the work organized, but you're doing the distributive property. \[(a + b)(c + d) = a(c+d) + b(c + d) = ac + ad + bc + bd\] Compare with \[(a+b)(c+d) = ac + ad + bc + bd\]Same thing.
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
Of course, if you prefer, this is equally valid: \[(a+b)(c+d) = ac + bc + ad + bd\] Whatever method gets you through it systematically and correctly....
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[A\star(B+ C)=A\star B+ A\star C\]
 one year ago
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