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

rebeccaskell94
 one year ago
Best ResponseYou've already chosen the best response.0Sure! 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

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

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

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

skullpatrol
 one year ago
Best 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.

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

Profunds77
 one year ago
Best ResponseYou've already chosen the best response.0I 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

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

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1FOIL 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.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1Of 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....

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