anonymous
  • anonymous
Perform the indicated operation. Be sure the answer is reduced.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1370976754120:dw|
anonymous
  • anonymous
@tcarroll010
ivettef365
  • ivettef365
sorry the other way k+ 5 -(k+2)(k-2) ----------- x ---------------- (k+5)(k-2) 7(k+2)

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anonymous
  • anonymous
\[\frac{ (k + 5)(4 - k ^{2}) }{ (k ^{2} + 3k - 10)(7k + 14) } = \frac{ \cancel{(k + 5)}(2 - k)\cancel{(2 + k)} }{ \cancel{(k + 5)}(k - 2)(7)\cancel{(k + 2)} } = \frac{ -1 }{ 7 }\]
ivettef365
  • ivettef365
now you can simplify what you have the same from the top and the bottom
anonymous
  • anonymous
The -1 in the numerator comes from: (k - 2) = (-1)(2 - k) in the denominator and the "2 - k" in the numerator and the denominator cancel as the last step.
anonymous
  • anonymous
All good now, @jumboabrfan ?
anonymous
  • anonymous
Where is 2-k in the original equation? (like where does it come from?)
anonymous
  • anonymous
In my first post, you can look at the factoring of: 4 - k^2 That's "2 - k" and "2 + k"
anonymous
  • anonymous
And what from the denominator is factored?
anonymous
  • anonymous
Because that's the difference of 2 perfect squares.
anonymous
  • anonymous
Do you see that now? @jumboabrfan ?
anonymous
  • anonymous
I see. I think I understand... thanks!! :)
anonymous
  • anonymous
uw! Good luck to you in all of your studies and thx for the recognition! @jumboabrfan
anonymous
  • anonymous
If you have any lingering question, please feel free to ask, and I'll be more than happy to help!

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