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Saylilbaby

  • one year ago

help pls im begging you all

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  1. anonymous
    • one year ago
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    with?

  2. pooja195
    • one year ago
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    It helps i you actually asked your question.

  3. pooja195
    • one year ago
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    *if

  4. saylilbaby
    • one year ago
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  5. saylilbaby
    • one year ago
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  6. saylilbaby
    • one year ago
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  7. saylilbaby
    • one year ago
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    @Michele_Laino

  8. saylilbaby
    • one year ago
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    @satellite73 @whpalmer4 @Zale101 @thomaster

  9. whpalmer4
    • one year ago
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    @saylilbaby are you here now?

  10. saylilbaby
    • one year ago
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    yes @whpalmer4

  11. saylilbaby
    • one year ago
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    @texaschic101

  12. whpalmer4
    • one year ago
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    Okay, let's look at the first one. Are you able to factor the numerator and denominator?

  13. whpalmer4
    • one year ago
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    \[\frac{n^4-11n^2+30}{n^4-7n^2+10}\]

  14. whpalmer4
    • one year ago
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    looking at the numerator, \(n^4-11n^2+30\) what are two numbers which when multiplied give you \(30\), but when added give you \(-11\)? Hopefully it is obvious that both must be negative numbers.

  15. saylilbaby
    • one year ago
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    is it C @whpalmer4 ?

  16. saylilbaby
    • one year ago
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    or A im confused @whpalmer4

  17. whpalmer4
    • one year ago
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    Let's work through the problem together, and you should be certain of your answer by the time we are done.

  18. saylilbaby
    • one year ago
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    i already worked thru i just needto state the restriction @whpalmer4

  19. whpalmer4
    • one year ago
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    Ok, didn't want to assume that...but it makes life easier :-) So, let's look at the factored denominator of the original expression: \[n^4-7n^2+10 = (n^2-5)(n^2-2)\]Agreed?

  20. whpalmer4
    • one year ago
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    When we find the restrictions, we have to do so on the unsimplified version, because we still have a "hole" in the function at those spots where we simplified away a term in the denominator. That means our restrictions on \(n\) here are located at all values of \(n\) where \[n^2-5=0\]and\[n^2-2=0\] Any value of \(n\) that makes either of those true will cause a division by \(0\) in the original expression.

  21. whpalmer4
    • one year ago
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    We can see that A is not the right answer, because the restrictions it imposes are \(n\ne5\) and \(n\ne2\), but those values do not cause the original denominator to be equal to \(0\). \[(5^2-5)(5^2-2) = 460\]\[(2^2-5)(2^2-2) = -2\]

  22. whpalmer4
    • one year ago
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    D is also not a correct answer for the same reason. \[(5^2-5)(5^2-2) = 460\]\[(-2)^2-5)((-2)^2-2) = -2\]

  23. whpalmer4
    • one year ago
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    B has the wrong fraction, so our hopes are left with C. Let's check it! \[n\ne \pm \sqrt{5}\]\[n\ne\pm\sqrt{2}\] If we evaluate the original denominator at \(n=\sqrt{5}\) \[(\sqrt{5}^2-5)(\sqrt{5}^2-2) = (5-5)(5-2)= 0 \] If we evaluate the original denominator at \(n=-\sqrt{5}\) \[((-\sqrt{5})^2-5)((-\sqrt{5})^2-2) = (5-5)(5-2)=0 \] So far, so good. Now trying \(n=\pm\sqrt{2}\) \[(\sqrt{2}^2-5)(\sqrt{2}^2-2) = (2-5)(2-2) = 0\] \[((-\sqrt{2})^2-5)((-\sqrt{2})^-2) = (2-5)(2-2) = 0\] Those are all good, so C is our answer.

  24. whpalmer4
    • one year ago
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    To find the restrictions, factor the initial denominator. Set each product term \(=0\) and find all the values which satisfy that. The list of all of such values for all of the terms is the set of restrictions on the expression.

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