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RH

  • 2 years 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
    • 2 years ago
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    Please help! :(

  2. DHASHNI
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    @amir.sat Sorry I don't understand :(

  5. amir.sat
    • 2 years ago
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    which part?

  6. RH
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Hmm... ok

  9. RH
    • 2 years ago
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    then what would be the right answer?

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

  11. RH
    • 2 years ago
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    Sorry, I am confused:(

  12. amir.sat
    • 2 years 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
    • 2 years ago
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    So the right answer would be A ?

  14. amir.sat
    • 2 years 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
    • 2 years ago
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    No prob. Thank you a lot!!! :)

  16. amir.sat
    • 2 years ago
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    ur welcome

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