anonymous
  • anonymous
What is NOT a restriction of the quotient of quantity 3 x squared plus 12 x plus 9 divided by quantity x squared minus 4 divided by quantity 4 x plus 4 divided by quantity x plus 2?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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e.mccormick
  • e.mccormick
Restrictions are generally invalid things, like negatives under a root or 0 on the botton of a fraction. So find out where the bottom would be 0 and that is the restriction.
anonymous
  • anonymous
how do i do that?
anonymous
  • anonymous
hllo?

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e.mccormick
  • e.mccormick
First, write it out with numbers, letters, and fractions sighs rather than words. Then, factor anything that would be the bottom of a fraction. If it is a 0, then it is restricted. \(\dfrac{{\dfrac{a}{b}}{\dfrac{c}{d}}\) In that: if b is 0, the top fraction is invalid; if d is 0, the bottom fraction is invalid; if c is 0, the bottom of the big fraction is invalid. So in that example, there are three things that can make it invalid. You can do something similar with your question once it is written out that way.
e.mccormick
  • e.mccormick
Oops... \(\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}\)
e.mccormick
  • e.mccormick
[(3x^2 + 12x + 9)/(x^2 - 4)]/[(4x + 4)/(x + 2)] So, any place (x^2 - 4), (4x + 4), or (x + 2) is = 0 is undefined. Each of those would make the bottom of the fraction into 0, which breaks rules.
anonymous
  • anonymous
\[x \neq1\] is that right ?
e.mccormick
  • e.mccormick
-1 is undefined, so restricted. THat is the 4x+4 one. 4(-1)+4 = ? -4+4 = ? 0 = 0
anonymous
  • anonymous
oh so it would be -1?
e.mccormick
  • e.mccormick
-1 is a restriction so it is not the answer.
e.mccormick
  • e.mccormick
If you find the three restrctions then you can pick what is left over as an answer... OR you can put in all the answers and see which one works.
e.mccormick
  • e.mccormick
Another way to check this sort of question is with a graphing tool. However, you need to know the math so that you can get the answer when there is no tool availble.

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