anonymous
  • anonymous
What is the simplified form of the equation below?
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
i look below i no see equation o.o
anonymous
  • anonymous
hold on!
anonymous
  • anonymous
HOLDING ON!! D:

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anonymous
  • anonymous
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anonymous
  • anonymous
dividing a fraction by a fraction is the same as multiplying a fraction by the reciprocal of another fraction...like this: \[\huge{{a\over b}\over{c\over d}}~~~~~\implies~~~~~{{a\over b}\times{d\over c}}\]
anonymous
  • anonymous
so in your case its:\[\huge{{12x\over x^2+6x+9}\over{4x^2\over x^2-9}}~~~~~\implies~~~~~{{12x\over x^2+6x+9}\times{x^2-9\over4x^2}}\] so now just cross multiply
anonymous
  • anonymous
here are the answers
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anonymous
  • anonymous
so it's the 4th one?
anonymous
  • anonymous
one sec :)
anonymous
  • anonymous
\[\large{{12x\over x^2+6x+9}\times{x^2-9\over4x^2}}~~~~~\implies~~~~{{12x\times(x^2-9)}\over{x^2+6x+9\times (4x^2)}}\]sorry i wrote that wrong...this gives you:\[\large{12x^3-108x\over 4x^4+24x^3+36x^2}\]simpllify that
anonymous
  • anonymous
i simplified it. it's the 2nd one
anonymous
  • anonymous
yup! :D
anonymous
  • anonymous
thnx, bro!
anonymous
  • anonymous
no problem!

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