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anonymous

  • 5 years ago

factor 225-(x+8)^2

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  1. anonymous
    • 5 years ago
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    Do you know the formula of the square difference? \[a^2-b^2=(a-b)(a+b)\]

  2. anonymous
    • 5 years ago
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    no i wasnt aware of that one

  3. anonymous
    • 5 years ago
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    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?

  4. anonymous
    • 5 years ago
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    (15-x+8)(15+X+8) ??

  5. anonymous
    • 5 years ago
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    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.

  6. anonymous
    • 5 years ago
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    HMM not sure i understand that one, thank you for the compliment btw

  7. anonymous
    • 5 years ago
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    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!

  8. anonymous
    • 5 years ago
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    ohhh i see it now witht the extra parens lol ok, haha and thanks again but your really way smarter man

  9. anonymous
    • 5 years ago
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    hey i have another one to ask if can help me still?

  10. anonymous
    • 5 years ago
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    Yeah sure, just one :)

  11. anonymous
    • 5 years ago
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    ok haha its Cos((pi/2)-x)/sin((pi/2)-x)

  12. anonymous
    • 5 years ago
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    \[\cos(\pi/2-x)\div \sin(\pi/2-x)\]

  13. anonymous
    • 5 years ago
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    i know i have to use the sum and difference id's for sin and cos but im hitting a wall

  14. anonymous
    • 5 years ago
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    You want to simplify: \[{\cos({\pi \over 2}-x) \over \sin({\pi \over 2}-x)}?\]

  15. anonymous
    • 5 years ago
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    yea the question says find

  16. anonymous
    • 5 years ago
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    It says find? A bit strange. Anyway, it clearly wants you to apply the following two formulas: \[\cos ({\pi \over 2}-x)=\sin x ..(1)\] \[\sin({\pi \over 2}-x)=\cos x .. (2)\]

  17. anonymous
    • 5 years ago
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    Are you following?

  18. anonymous
    • 5 years ago
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    i think, would it come out to tangent?

  19. anonymous
    • 5 years ago
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    Exactly!! Didn't I just I say you're smart?!

  20. anonymous
    • 5 years ago
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    haha i try

  21. anonymous
    • 5 years ago
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    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.

  22. anonymous
    • 5 years ago
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    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!!

  23. anonymous
    • 5 years ago
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    You're welcome. I am sure you're very good at math, and you pick things quickly. Good luck!!

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