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RH

  • one year ago

When solving the polynomial inequality 4x^5 - 18x^4 + 18x^3 ≥ 0 how many graphs will you draw before you draw your final graph? A. two B. four C. five D. six

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  1. RH
    • one year ago
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    Please help! :(

  2. DHASHNI
    • one year ago
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    http://www.wolframalpha.com/input/?i=4x%5E5+-+18x%5E4+%2B+18x%5E3+%E2%89%A5+0&dataset=

  3. amir.sat
    • one year ago
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    4x^5 - 18x^4 + 18x^3 ≥ 0 divide it by x^2 so it's 4x^3 - 18x^2 + 18x ≥ 0 for the x>0 it's 4x^2 - 18x + 18 ≥ 0 (first graph) and for the x <0 it's 4x^2 - 18x + 18 < 0 (second graph) if you draw the first graph the the impossible rations for x>0 is \[\left( 0 , 1.5 \right) \cup \left( 3 , \infty \right) \cup \ { 0 , 1.5 , 3}\] but the second graph for the x<0 doesn't have any rations so the total answer is \[\left( 0 , 1.5 \right) \cup \left( 3 , \infty \right) \cup \ { 0 , 1.5 , 3}\]

  4. RH
    • one year ago
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    @amir.sat Sorry I don't understand :(

  5. amir.sat
    • one year ago
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    which part?

  6. RH
    • one year ago
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    Now I understand, so when solving the inequality, 2 graphs will be drawn before the final graph?

  7. amir.sat
    • one year ago
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    the point is that you divide an inequality by variable that can be negative you should separate it to two inequality one for the negative variable and one for the positive variable

  8. RH
    • one year ago
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    Hmm... ok

  9. RH
    • one year ago
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    then what would be the right answer?

  10. amir.sat
    • one year ago
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    all of 'em i meant the sum of each seperated graph's answer

  11. RH
    • one year ago
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    Sorry, I am confused:(

  12. amir.sat
    • one year ago
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    you separated the inequality into two inequality when you divide it by a variable i meant the answer is the sum of the separated inequality's answer

  13. RH
    • one year ago
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    So the right answer would be A ?

  14. amir.sat
    • one year ago
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    oh sorry .yes it's A i didn't see the options , i just thought it's just the questions

  15. RH
    • one year ago
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    No prob. Thank you a lot!!! :)

  16. amir.sat
    • one year ago
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    ur welcome

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