anonymous
  • anonymous
help!? pleaseeee :) Simplify (6x2+ 11x – 3) + (2x2– 17x – 4) 8x2+ 6x – 1 8x2– 6x – 7 8x2+ 6x + 1 8x2– 6x + 7 How do you simplify these equations?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
add like terms
raden_zaikaria
  • raden_zaikaria
6x2+ 11x – 3 + 2x2– 17x – 4
anonymous
  • anonymous
(6x2+ 11x – 3) + (2x2 – 17x – 4) 8x^2 -4x -7

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More answers

anonymous
  • anonymous
\[6x^2+ 11x – 3 + 2x^2– 17x – 4\]
anonymous
  • anonymous
B I think
anonymous
  • anonymous
Yeah Its B
anonymous
  • anonymous
Setting it up like this makes it alot easier: (6x2+ 11x – 3) (2x2– 17x – 4)
anonymous
  • anonymous
6x2 + 2x2 = 8x2 11x - 17x = -6x -3 - 4 = -7
raden_zaikaria
  • raden_zaikaria
6x2+ 2x2=8x2 and 11x – 17x = -6x and – 3 – 4 = -7
anonymous
  • anonymous
(6x2+ 11x – 3) + (2x2 – 17x – 4) 8x^2 -6x -7 sorry
raden_zaikaria
  • raden_zaikaria
that ansmwer is 8x2-6x-7
anonymous
  • anonymous
Thank you! everyone! :) I get it now :)
anonymous
  • anonymous
Any time
raden_zaikaria
  • raden_zaikaria
welcome from malaysia hehe
anonymous
  • anonymous
Simplify 2x2(4x2+ 9x – 13) 6x4+ 18x3– 26x2 8x4+ 18x3– 26x2 8x2+ 18x – 26 6x2+ 18x – 26
anonymous
  • anonymous
would the answer be D? or A...
anonymous
  • anonymous
none
anonymous
  • anonymous
oh........ why?
anonymous
  • anonymous
\[2x^2(4x^2+ 9x – 13)\] now we are multiplying
anonymous
  • anonymous
so \[2x^2\] should multiply each item in the bracket
anonymous
  • anonymous
hmm, so C?
anonymous
  • anonymous
almost there but notice that even the power changes \[2x^2(4x^2)=8x^{4}\]
anonymous
  • anonymous
the powers add each other\[x^a(x^b)=x^{a+b}\]
anonymous
  • anonymous
oh. ok. I wasnt sure if i was supposed to change the powers as well.
anonymous
  • anonymous
Which of the following represents the area of a rectangle whose length is x – 7 and whose width is x + 12? x2+ 5x – 84 x2– 84 x2– 5x – 84 x2– 84x + 5 IS C THE CORRECT ANSWER?
anonymous
  • anonymous
lol...
anonymous
  • anonymous
what?
anonymous
  • anonymous
so did you get the answer to the previous one
anonymous
  • anonymous
Yes I did.
anonymous
  • anonymous
you shud ask a new question as a new question
anonymous
  • anonymous
I'm not 100% sure on this question :l
anonymous
  • anonymous
ok,
anonymous
  • anonymous
I'll close this question and post a new one so i can give u a medal.

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