anonymous
  • anonymous
What is the simplified form of the quantity y-squared plus 7y plus 12 over the quantity y-squared minus 3y minus 18 ? Answers: y minus 4 over y minus 6 y minus 4 over y plus 6 y plus 4 over y minus 6 y plus 4 over y plus 6
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
\(\Large y^2 + 7y + 12\) factors to what?
anonymous
  • anonymous
(Y+3)(Y+4)?
jim_thompson5910
  • jim_thompson5910
correct

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
\(\Large y^2 - 3y - 18\) factors to what?
anonymous
  • anonymous
(y-6)(y+3) not sure about that one
jim_thompson5910
  • jim_thompson5910
it's correct
anonymous
  • anonymous
ok, so what do i do next? cancel out (y+3)?
jim_thompson5910
  • jim_thompson5910
Why did we do all that? We factor everything, then we cancel the common (y+3) terms \[\Large \frac{y^2+7y+12}{y^2-3y-18}\] \[\Large \frac{(y+3)(y+4)}{y^2-3y-18}\] \[\Large \frac{(y+3)(y+4)}{(y-6)(y+3)}\] \[\Large \frac{\color{red}{(y+3)}(y+4)}{(y-6)\color{red}{(y+3)}}\] \[\Large \frac{\color{red}{\cancel{(y+3)}}(y+4)}{(y-6)\color{red}{\cancel{(y+3)}}}\] \[\Large \frac{y+4}{y-6}\] ------------------------------------------------------- So \(\Large \frac{y^2+7y+12}{y^2-3y-18}\) simplifies to \(\Large \frac{y+4}{y-6}\)
jim_thompson5910
  • jim_thompson5910
yes that's correct
anonymous
  • anonymous
thank you so much!!!!
jim_thompson5910
  • jim_thompson5910
you're welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.