anonymous
  • anonymous
Simplify each of the following expressions as much as possible..
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
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anonymous
  • anonymous
\[\frac{ (i+2)! }{ (i-1)! }\]
anonymous
  • anonymous
Not sure how to start these..explain?
anonymous
  • anonymous
they dont like i in the denominator multiply by complex conjugate :)

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anonymous
  • anonymous
well madam,(i+2)! = (i+2)(i+1)(i)... and (i-1)! = (i-1)(i-2)... and do you have any idea to solve it? :)
anonymous
  • anonymous
need help?
anonymous
  • anonymous
Does something cancel out along the way?
anonymous
  • anonymous
no :)
anonymous
  • anonymous
I'm all confused! How do you know what to do?
anonymous
  • anonymous
ok,i will tell you: notice this examples,and try it yourself... \[\frac{ (5+2)(5+1)(5)(5-1)(5-2)(5-3)(5-4) }{ (5-1)(5-2)(5-3)(5-4) }\]
anonymous
  • anonymous
and can you solve it?
anonymous
  • anonymous
I just don't understand what I'm trying to do. I can see that you're using the same pattern and that after a certain point you're getting the same numbers
anonymous
  • anonymous
you should solve it like this:
1 Attachment
anonymous
  • anonymous
so the answer is 7x6x5 = 210
anonymous
  • anonymous
So it does kind of like cancel out? Would my answer just be i+2?
anonymous
  • anonymous
no the answer won't be i+2 , see: \[\frac{ (i+2)(i+1)(i)(i-1)... }{ (i-1)(i-2)... }\]
anonymous
  • anonymous
i-1? I don't get how it'll stop at i, after you cross out i-1.
anonymous
  • anonymous
I can't wrap my mind around it I'm sorry :(
anonymous
  • anonymous
we have i-1 , i-2 , i-3 , ...so we should cross out i-1 , i-2 , i -3 , ... and WE CAN'T CROSS OUT i+2 , i+1 , i so the answer is i+2 , i+1 , i
anonymous
  • anonymous
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anonymous
  • anonymous
But that's it??????
anonymous
  • anonymous
what?
anonymous
  • anonymous
Okay that makes sense I think. Thank you
anonymous
  • anonymous
welcome madam :)

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