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anonymous

  • 5 years ago

can anyone help me understand this problem Solve using the addition principle. Graph and write set-builder 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 {t|t ≥ 0} or (–∞, 0]

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  1. anonymous
    • 5 years ago
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    what class is it for. Algebra or set theory?

  2. anonymous
    • 5 years ago
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    basic algebra

  3. anonymous
    • 5 years ago
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    Can you tell me how you arrived at that solution?

  4. anonymous
    • 5 years ago
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    5(t + 3) + 9 ≥ 3(t − 2) − 10 5t+3+9 ≥ 3t-2-10 5t+12 ≥ 3t-12 5t-3t +12-12 ≥ 3t-3t-12+12 2t /2 ≥ 0/2 t≥0 that was the answer i got

  5. anonymous
    • 5 years ago
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    I think I see what you did wrong

  6. anonymous
    • 5 years ago
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    5(t+3)+9>or= 3(t-2)-10 You have to use distributive property.(Everything inside the parenthesis should be multiplied by number outside.)

  7. anonymous
    • 5 years ago
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    dam it cant believe i let that one by me

  8. anonymous
    • 5 years ago
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    So it is 5*t+5*3+9>0r=3*t+(3*-2) -10

  9. anonymous
    • 5 years ago
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    So did you end up getting an answer?

  10. anonymous
    • 5 years ago
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    working it out now

  11. anonymous
    • 5 years ago
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    5(t + 3) + 9 ≥ 3(t − 2) − 10 5t+15+9 ≥ 3t-6-10 5t+24 ≥ 3t-16 5t-3t +24-24 ≥ 3t-3t-16+24 2t /2 ≥ 8/2 t≥4 {t|t ≥ 4} or (–∞, 4]

  12. anonymous
    • 5 years ago
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    not sure about the last line though thats whats messing me up

  13. anonymous
    • 5 years ago
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    since t is greater or equal to 4 we should start from 4 to infinity. so it should be [4,infinity)

  14. anonymous
    • 5 years ago
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    ok so i have the answer backwards 5(t + 3) + 9 ≥ 3(t − 2) − 10 5t+15+9 ≥ 3t-6-10 5t+24 ≥ 3t-16 5t-3t +24-24 ≥ 3t-3t-16+24 2t /2 ≥ 8/2 t≥4 {t|t ≥ 4} or [4,∞)

  15. anonymous
    • 5 years ago
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    thank you i see im going to hate my next math class more then this one

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