anonymous
  • anonymous
Help! PLEASE :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
It's ok :( lol
anonymous
  • anonymous
well,can you say this: (x+y)^2 = ?

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anonymous
  • anonymous
need help?
anonymous
  • anonymous
I do need help lol
anonymous
  • anonymous
(x+y)^2 = (x+y)(x+y) : |dw:1378920609376:dw|
anonymous
  • anonymous
can you do it now?
anonymous
  • anonymous
Hang on just a moment so I can open the quiz back up please
anonymous
  • anonymous
I'm still a little confused
anonymous
  • anonymous
Well can you tell me if I'm correct. I think B and C are correct
anonymous
  • anonymous
wait,you will give the answer...
anonymous
  • anonymous
ok ,ok , I will prove it :) (x+y)(x+y) = x^2 + xy + y^2 + xy = x^2 + 2xy + y^2
anonymous
  • anonymous
did you understand it?
anonymous
  • anonymous
Kind of... so is b and c the right answers? or are there more?
anonymous
  • anonymous
well, was (x+y)^2 equal to x^2 + y^2 ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
it was?????? I proved it just a few moments ago! (x+y)^2=(x+y)(x+y) = x^2 + xy + y^2 + xy = x^2 + 2xy + y^2
anonymous
  • anonymous
so it's x^2 + 2xy + y^2 not x^2 + y^2
anonymous
  • anonymous
Ohhhhhhhhhh Ahh lord! ok ok yes I get it now
anonymous
  • anonymous
Which means it's not b either
anonymous
  • anonymous
Correct?
anonymous
  • anonymous
yes,(x+y)^2 was not equal to x^2 + y^2 so (x-y)^3 is not x^3 - y^3
anonymous
  • anonymous
D seems correct
anonymous
  • anonymous
D and A is correct,do you want the proof?
anonymous
  • anonymous
YES!
anonymous
  • anonymous
\[(x-y)^2 = (x-y)(x-y) = x^2 - xy - xy + y^2 = x^2 - 2xy + y^2\]
anonymous
  • anonymous
and
anonymous
  • anonymous
\[(x-y)^3 = (x-y)^2 \times (x-y)\]
anonymous
  • anonymous
ok?
anonymous
  • anonymous
now can you prove the second one?
anonymous
  • anonymous
I get it now. Much appreciation! Like, so much lol
anonymous
  • anonymous
welcome :)

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