anonymous
  • anonymous
factor 225-(x+8)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Do you know the formula of the square difference? \[a^2-b^2=(a-b)(a+b)\]
anonymous
  • anonymous
no i wasnt aware of that one
anonymous
  • anonymous
Well you should use this formula, you can rewrite your expression as: \[15^2-(x+8)^2\] Now you have a=15, and b=x+8. Tell me what you think it would look like after factorization?

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anonymous
  • anonymous
(15-x+8)(15+X+8) ??
anonymous
  • anonymous
You're smart. You just made a little mistake with a sign. Deal with x+8 as a combination at first. So when you subtract, a minus sign will go to both x and 8.
anonymous
  • anonymous
HMM not sure i understand that one, thank you for the compliment btw
anonymous
  • anonymous
Well, see here: \[15^2-(x+8)^2=(15-(x+8))(15+(x+8))=(15-x-8)(15+x+8)\] \[=(7-x)(23+x)\] And yes you really are good!
anonymous
  • anonymous
ohhh i see it now witht the extra parens lol ok, haha and thanks again but your really way smarter man
anonymous
  • anonymous
hey i have another one to ask if can help me still?
anonymous
  • anonymous
Yeah sure, just one :)
anonymous
  • anonymous
ok haha its Cos((pi/2)-x)/sin((pi/2)-x)
anonymous
  • anonymous
\[\cos(\pi/2-x)\div \sin(\pi/2-x)\]
anonymous
  • anonymous
i know i have to use the sum and difference id's for sin and cos but im hitting a wall
anonymous
  • anonymous
You want to simplify: \[{\cos({\pi \over 2}-x) \over \sin({\pi \over 2}-x)}?\]
anonymous
  • anonymous
yea the question says find
anonymous
  • anonymous
It says find? A bit strange. Anyway, it clearly wants you to apply the following two formulas: \[\cos ({\pi \over 2}-x)=\sin x ..(1)\] \[\sin({\pi \over 2}-x)=\cos x .. (2)\]
anonymous
  • anonymous
Are you following?
anonymous
  • anonymous
i think, would it come out to tangent?
anonymous
  • anonymous
Exactly!! Didn't I just I say you're smart?!
anonymous
  • anonymous
haha i try
anonymous
  • anonymous
So, applying the two formulas gives: \[{\sin x \over \cos x}=\tan x\] I don't know if it's giving any values for x.
anonymous
  • anonymous
no it was just asking for an equal trig function, i didnt know those equations though, putting them in my notes, thanks man, youve helped me alot!!
anonymous
  • anonymous
You're welcome. I am sure you're very good at math, and you pick things quickly. Good luck!!

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