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anonymous

  • one year ago

Help please this is my last year and i really need to do good! I will fan and medal!!! Given f(x)=2x^2+8 and g(x)=5x-6, find each function for questions 4-5. 4. (f+g)(x) 5. (fxg)(x)

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  1. breyliance
    • one year ago
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    i would suggest contacting your teacher

  2. anonymous
    • one year ago
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    I cant he is out of the office today

  3. breyliance
    • one year ago
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    ask a parent interweb idk

  4. anonymous
    • one year ago
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    \[(f+g)(x)=f(x) + g(x)\] Add the functions and combine like terms \[f(x) + g(x)=2x^2+8+5x-6\]

  5. breyliance
    • one year ago
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    holy crap

  6. anonymous
    • one year ago
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    :( none of my family knows this stuff and either do i, i have trouble understanding the problems

  7. anonymous
    • one year ago
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    5. \((f \times g)(x)=f(x) \times g(x)=(2x^2+8)(5x-6)\) Use FOIL/Distribution to simplify

  8. skullpatrol
    • one year ago
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    |dw:1441200012218:dw|

  9. skullpatrol
    • one year ago
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    To add polynomials you add their similar terms, similar terms monomials that are exactly alike or are the same except for their numerical coefficient.

  10. skullpatrol
    • one year ago
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    |dw:1441200483316:dw|

  11. anonymous
    • one year ago
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    So nothing else gets done with the 2x^2 or the 5x?

  12. skullpatrol
    • one year ago
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    Correct, they are not similar terms.

  13. anonymous
    • one year ago
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    But if there wasnt the ^2 then they could have been added together to make 7x right?

  14. skullpatrol
    • one year ago
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    yes

  15. anonymous
    • one year ago
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    okay how would i work the next one out?

  16. skullpatrol
    • one year ago
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    Use the distributive property. Do you recall that?

  17. skullpatrol
    • one year ago
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    |dw:1441201094779:dw|

  18. skullpatrol
    • one year ago
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    |dw:1441201244829:dw|

  19. skullpatrol
    • one year ago
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    |dw:1441201528521:dw|

  20. anonymous
    • one year ago
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    Thank you so much, Would you be able to help me with a few more in a sec?

  21. skullpatrol
    • one year ago
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    I can try :-)

  22. anonymous
    • one year ago
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    Okay ill post a new question though

  23. skullpatrol
    • one year ago
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    sure

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