countrygirl1431
  • countrygirl1431
Please help Algebra 2
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
countrygirl1431
  • countrygirl1431
Perform the indicated operation.
countrygirl1431
  • countrygirl1431
https://alphaomegaca.ignitiaschools.com/media/g_alg02_2015/5/184.gif
iGreen
  • iGreen
Flip the 2nd fraction and multiply

Looking for something else?

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

More answers

countrygirl1431
  • countrygirl1431
@iGreen
countrygirl1431
  • countrygirl1431
\[\frac{ (a+3)^2 }{ a-3 }\times \frac{ a^2-9 }{5 }\]
iGreen
  • iGreen
Correct, and can you multiply those 2 together?
iGreen
  • iGreen
Multiply the numerators together and multiply the denominators together.
countrygirl1431
  • countrygirl1431
im not sure how
countrygirl1431
  • countrygirl1431
i know it starts off \[2a^2\]
iGreen
  • iGreen
Okay, for the numerator we have \(\sf (a + 3)(a + 3)(a^2 - 9)\)
iGreen
  • iGreen
Can you multiply \(\sf (a + 3)(a + 3)\)?
countrygirl1431
  • countrygirl1431
a+9
iGreen
  • iGreen
No.. \(\sf a \times a\) \(\sf a \times 3\) \(\sf 3 \times a\) \(\sf 3 \times 3\) Multiply those
countrygirl1431
  • countrygirl1431
im confused
iGreen
  • iGreen
\(\sf a \times a \rightarrow a^2\) \(\sf a \times 3 \rightarrow 3a\) \(\sf 3 \times a \rightarrow 3a\) \(\sf 3 \times 3 \rightarrow 9\)
iGreen
  • iGreen
Which gives us \(\sf a^2 + 6a + 9\), now multiply that to \(\sf a^2 - 9\)
countrygirl1431
  • countrygirl1431
ok i thought tht was it but i wasnt sure
countrygirl1431
  • countrygirl1431
so would this be a^4
countrygirl1431
  • countrygirl1431
or 6a^5
iGreen
  • iGreen
You gotta multiply all of them
iGreen
  • iGreen
\(\sf a^2 \times a^2\) \(\sf a^2 \times -9\) \(\sf 6a \times a^2\) \(\sf 6a \times -9\) \(\sf 9 \times a^2\) \(\sf 9 \times -9\)
iGreen
  • iGreen
The first one equals \(\sf a^4\), that's correct.
countrygirl1431
  • countrygirl1431
the 2nd 1 is a^2-9
iGreen
  • iGreen
No, it's \(\sf -9a^2\)
countrygirl1431
  • countrygirl1431
oh ok
countrygirl1431
  • countrygirl1431
3=6a^3
iGreen
  • iGreen
Correct.
countrygirl1431
  • countrygirl1431
would 4 be -54a
countrygirl1431
  • countrygirl1431
5 wouldbt 9a^2 and 6 would be -81
iGreen
  • iGreen
4 is correct.
iGreen
  • iGreen
So are 5 and 6.
countrygirl1431
  • countrygirl1431
ok
iGreen
  • iGreen
That gives us: \(\sf a^4 - 9a^2 + 6a^3 - 54a + 9a^2 - 81\)
iGreen
  • iGreen
Combine like terms and you got: \(\sf a^4 + 6a^3 - 54a - 81\)
iGreen
  • iGreen
Now simplify the denominator. \(\sf 5(a - 3)\)
countrygirl1431
  • countrygirl1431
5a-3
iGreen
  • iGreen
Multiply 5 to both terms.
iGreen
  • iGreen
\(\sf 5 \times a\) \(\sf 5 \times -3\)
countrygirl1431
  • countrygirl1431
5a and -15
iGreen
  • iGreen
Yes, that gives us: \(\sf\dfrac{ a^4 + 6a^3 - 54a - 81}{5a - 15}
countrygirl1431
  • countrygirl1431
i cant read it @green_1
countrygirl1431
  • countrygirl1431
@iGreen
countrygirl1431
  • countrygirl1431
sorry @green_1 i meant to tag someone else
green_1
  • green_1
you mean the equation above
countrygirl1431
  • countrygirl1431
the last thing posted i cant read it
countrygirl1431
  • countrygirl1431
@iGreen could u repost the last thing

Looking for something else?

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