Saylilbaby
  • Saylilbaby
help pls im begging you all
Mathematics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
with?
pooja195
  • pooja195
It helps i you actually asked your question.
pooja195
  • pooja195
*if

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Saylilbaby
  • Saylilbaby
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whpalmer4
  • whpalmer4
@saylilbaby are you here now?
Saylilbaby
  • Saylilbaby
Saylilbaby
  • Saylilbaby
whpalmer4
  • whpalmer4
Okay, let's look at the first one. Are you able to factor the numerator and denominator?
whpalmer4
  • whpalmer4
\[\frac{n^4-11n^2+30}{n^4-7n^2+10}\]
whpalmer4
  • whpalmer4
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.
Saylilbaby
  • Saylilbaby
is it C @whpalmer4 ?
Saylilbaby
  • Saylilbaby
or A im confused @whpalmer4
whpalmer4
  • whpalmer4
Let's work through the problem together, and you should be certain of your answer by the time we are done.
Saylilbaby
  • Saylilbaby
i already worked thru i just needto state the restriction @whpalmer4
whpalmer4
  • whpalmer4
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?
whpalmer4
  • whpalmer4
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.
whpalmer4
  • whpalmer4
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 \(0\). \[(5^2-5)(5^2-2) = 460\]\[(2^2-5)(2^2-2) = -2\]
whpalmer4
  • whpalmer4
D is also not a correct answer for the same reason. \[(5^2-5)(5^2-2) = 460\]\[(-2)^2-5)((-2)^2-2) = -2\]
whpalmer4
  • whpalmer4
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.
whpalmer4
  • whpalmer4
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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