Saylilbaby
  • Saylilbaby
help pls im begging you all
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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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Saylilbaby
  • Saylilbaby
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Saylilbaby
  • Saylilbaby
@Michele_Laino
Saylilbaby
  • Saylilbaby
@satellite73 @whpalmer4 @Zale101 @thomaster
whpalmer4
  • whpalmer4
@saylilbaby are you here now?
Saylilbaby
  • Saylilbaby
yes @whpalmer4
Saylilbaby
  • Saylilbaby
@texaschic101
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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