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greeneyes<3

Find the domain of the expression: 1. 6x^2-9, x>0 2. (x^2-x-12)/(x^2-8+16) Please show steps or explain them to me! Thank you!

  • 7 months ago
  • 7 months ago

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  1. munaali
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    -9

    • 7 months ago
  2. greeneyes<3
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    is it like all real numbers, or can you please explain?

    • 7 months ago
  3. munaali
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    yes

    • 7 months ago
  4. greeneyes<3
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    it still does not make sense, x has to be > 0

    • 7 months ago
  5. munaali
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    the real numer has negive

    • 7 months ago
  6. greeneyes<3
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    @jim_thompson5910 can you help me with these two problems?

    • 7 months ago
  7. jim_thompson5910
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    by default, all polynomials have a domain of the set of all real numbers

    • 7 months ago
  8. jim_thompson5910
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    but because it clearly states that x > 0, that means the domain is x > 0, so the domain is the set of positive real numbers

    • 7 months ago
  9. jim_thompson5910
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    to find the domain of (x^2-x-12)/(x^2-8x+16), set the denominator equal to zero and solve for x those two solutions will be values you exclude from the domain

    • 7 months ago
  10. greeneyes<3
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    Thank you! Now my answers are: 1. x=all + real numbers 2. x cannot=+ or - 4, and -2 is that right?

    • 7 months ago
  11. jim_thompson5910
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    the first one is correct, but the second is not

    • 7 months ago
  12. jim_thompson5910
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    hint: x^2 - 8x + 16 factors to (x-4)(x-4)

    • 7 months ago
  13. greeneyes<3
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    so x cannot equal 4, or -2 ?

    • 7 months ago
  14. jim_thompson5910
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    how are you getting -2?

    • 7 months ago
  15. greeneyes<3
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    oh nevermind I see what I was doing wrong, was separating x^2 and -8x and solving them each for -16 separately. So the answer is just x cannot = 4 ?

    • 7 months ago
  16. jim_thompson5910
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    bingo, the domain is the set of all real numbers but x cannot equal 4

    • 7 months ago
  17. greeneyes<3
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    Okay thank you! :)

    • 7 months ago
  18. jim_thompson5910
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    you're welcome

    • 7 months ago
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