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justin890 Group TitleBest ResponseYou've already chosen the best response.0
i think i asked this before and it got answerd..but i ruined my hw and lost the answer..:(
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
that solution was correct ^^
 one year ago

justin890 Group TitleBest ResponseYou've already chosen the best response.0
what one?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
in ur figure, all steps have done
 one year ago

justin890 Group TitleBest ResponseYou've already chosen the best response.0
i know..but i need to justify each step..i dont understand this question at all..
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
well, this is a algebraic problem... given the length of AC=32, from the figure the length of AB=2x and BC=2x+8. doesnt AC=AB+BC ? just to add both lines, to make an expression algebraic and solve for x. its solution like in figure :)
 one year ago

justin890 Group TitleBest ResponseYou've already chosen the best response.0
my brain hurts i cant process ANY of that..could you make that easier to understand?
 one year ago

justin890 Group TitleBest ResponseYou've already chosen the best response.0
could you do it step by step?
 one year ago

kelly226 Group TitleBest ResponseYou've already chosen the best response.5
1) Segment addition and definition of between 2)substitution with given information 3) simplifying 2x + 6x = x(2 + 6) = 8x , it sorta an distribute thing 4)subtraction property of equality ( subtracted 8 both sides) 5)division property of equality ( multiplicative inverse property) (divide by 8 on both sides)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
AB+BC=AC, put the values of them : 2x+6x+8=32, simplify it 8x+8=32, add 8 from both sides 8x+88=328,simplify it 8x=24, divide both sides by 8 8x/8 = 24/8, solve for x x=3
 one year ago

justin890 Group TitleBest ResponseYou've already chosen the best response.0
thank you @RadEn :)
 one year ago
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