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anonymous
 5 years ago
can anyone help me understand this problem Solve using the addition principle. Graph and write setbuilder notation and interval notation for each answer.
5(t + 3) + 9 ≥ 3(t − 2) − 10 i have solved it but not sure if i got it right
{tt ≥ 0} or (–∞, 0]
anonymous
 5 years ago
can anyone help me understand this problem Solve using the addition principle. Graph and write setbuilder notation and interval notation for each answer. 5(t + 3) + 9 ≥ 3(t − 2) − 10 i have solved it but not sure if i got it right {tt ≥ 0} or (–∞, 0]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what class is it for. Algebra or set theory?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you tell me how you arrived at that solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.05(t + 3) + 9 ≥ 3(t − 2) − 10 5t+3+9 ≥ 3t210 5t+12 ≥ 3t12 5t3t +1212 ≥ 3t3t12+12 2t /2 ≥ 0/2 t≥0 that was the answer i got

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I see what you did wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.05(t+3)+9>or= 3(t2)10 You have to use distributive property.(Everything inside the parenthesis should be multiplied by number outside.)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dam it cant believe i let that one by me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it is 5*t+5*3+9>0r=3*t+(3*2) 10

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So did you end up getting an answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.05(t + 3) + 9 ≥ 3(t − 2) − 10 5t+15+9 ≥ 3t610 5t+24 ≥ 3t16 5t3t +2424 ≥ 3t3t16+24 2t /2 ≥ 8/2 t≥4 {tt ≥ 4} or (–∞, 4]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not sure about the last line though thats whats messing me up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since t is greater or equal to 4 we should start from 4 to infinity. so it should be [4,infinity)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so i have the answer backwards 5(t + 3) + 9 ≥ 3(t − 2) − 10 5t+15+9 ≥ 3t610 5t+24 ≥ 3t16 5t3t +2424 ≥ 3t3t16+24 2t /2 ≥ 8/2 t≥4 {tt ≥ 4} or [4,∞)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you i see im going to hate my next math class more then this one
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