anonymous
  • anonymous
.
Mathematics
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anonymous
  • anonymous
.
Mathematics
katieb
  • katieb
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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.

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anonymous
  • anonymous
anonymous
  • anonymous
could you please do both? Thanks
rhr12
  • rhr12
sorry for the wrong try for last question. Not enough brave to try another one.

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anonymous
  • anonymous
can someone please help?
anonymous
  • anonymous
anonymous
  • anonymous
The last question is D. but would you want me to explain how to get that?
anonymous
  • anonymous
Did you need the first question too?
anonymous
  • anonymous
anonymous
  • anonymous
Scared bro. @rhr12
anonymous
  • anonymous
yes
anonymous
  • anonymous
Yes to both questions?
anonymous
  • anonymous
yes
anonymous
  • anonymous
think the first one is A
anonymous
  • anonymous
Question 6:\[\frac{ x^2-15x+56 }{ x-7 }\]
anonymous
  • anonymous
Yes that's what I thought too.
anonymous
  • anonymous
\[\frac{ (x-7)(x-8) }{ x-7 }\]
anonymous
  • anonymous
Do you know how I got to that step?
anonymous
  • anonymous
yes
anonymous
  • anonymous
|dw:1438449674703:dw|
anonymous
  • anonymous
what about this? Simplify completely quantity 12 x plus 36 over quantity x squared minus 4 x minus 21 and find the restrictions on the variable.
anonymous
  • anonymous
Cancel the (x-7) and you get (x-8) which is D.
anonymous
  • anonymous
Can you show that as an equation or picture? It's easier to understand that way. :)
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
|dw:1438449862391:dw|
anonymous
  • anonymous
Answer A.
anonymous
  • anonymous
It can't be positive 7 since the denominator will turn 0 making the answer not available or 0.
anonymous
  • anonymous
Did you get that?
anonymous
  • anonymous
anonymous
  • anonymous
What do you mean?
anonymous
  • anonymous
abcd
anonymous
  • anonymous
A
anonymous
  • anonymous
Are they the same kind of questions?
anonymous
  • anonymous
Let me see them.
anonymous
  • anonymous
anonymous
  • anonymous
Okay hold on.
anonymous
  • anonymous
Ya I am
anonymous
  • anonymous
okay i gtg just do all of the steps for the problems and i'll look at them when i come back
anonymous
  • anonymous
Alright
anonymous
  • anonymous
A. for the question above Question 9.\[\frac{ 5x^2-21x-20 }{ 5x^2-16x-16 }\]\[\frac{ (x-5)(5x+4) }{ (x-4)(5x+4) }\]\[\frac{ (x-5) }{ (x-4) }\]
anonymous
  • anonymous
Question 9:\[\frac{ 2x^2+20x+32 }{ x^2-2x-80 }\]\[\frac{ 2(x^2+10x+16) }{ x^2-2x-80 }\]\[\frac{ 2(x+8)(x+2) }{ (x-10)(x+8) }\]\[\frac{ 2(x+2) }{ (x-10) }\]
anonymous
  • anonymous
\[\frac{ 2x^2+16x+30 }{ 5x^2+13x-6 }\]\[\frac{ 2(x^2+8x+15) }{ 5x^2+13x-6 }\]\[\frac{ 2(x+3)(x+5) }{ (x+3)(5x-2) }\]\[\frac{ 2(x+5) }{ (5x-2) }\]This is the question that is by itself in the picture. Hope this helps! :)

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