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james1769

  • one year ago

Add (3x4 − 2x3 + 1) + (12x4 + x2 − 11). (1 point) 15x4 − x3 − 10 15x4 − x2 − 10 15x4 − 2x3 + x2 + 10 15x4 − 2x3 + x2 − 10

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  1. the_ocean_girl
    • one year ago
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    need a walk through?

  2. james1769
    • one year ago
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    yes please

  3. james1769
    • one year ago
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    anyone?

  4. the_ocean_girl
    • one year ago
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    sorry i was taking notes

  5. the_ocean_girl
    • one year ago
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    i'm here to help walk you through

  6. james1769
    • one year ago
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    thnks

  7. the_ocean_girl
    • one year ago
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    yup okay so first, do the multiplication inside the parentheses, but remember, the 2 in 2x3 is negative

  8. james1769
    • one year ago
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    ok

  9. the_ocean_girl
    • one year ago
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    so after you do that, you combine all the numbers in the parentheses then you add the sums that are in the 2 sets of parentheses together do you understand it now?

  10. james1769
    • one year ago
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    kinda math isnt my friend

  11. the_ocean_girl
    • one year ago
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    math is nobody's friend

  12. james1769
    • one year ago
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    nope

  13. james1769
    • one year ago
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    so the answer is d

  14. the_ocean_girl
    • one year ago
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    yup!

  15. james1769
    • one year ago
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    thanks

  16. the_ocean_girl
    • one year ago
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    no problem!

  17. james1769
    • one year ago
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    i have a few more if u want to help

  18. the_ocean_girl
    • one year ago
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    sure

  19. james1769
    • one year ago
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    Subtract (5x2 + 3) −(2x2 + 4x − 12).

  20. james1769
    • one year ago
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    3x2 − 4x + 15 3x2 + 4x + 15 3x2 − 4x − 9 3x2 + 4x − 9

  21. the_ocean_girl
    • one year ago
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    okay so same thing as before, but this time, you need to distribute the negative that's outside of the second set of parentheses basically, because of that negative, you have to multiply everything inside of that 2nd set of parentheses by -1 before anything else

  22. james1769
    • one year ago
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    do u add the exponents this is part of a dif qu

  23. james1769
    • one year ago
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    @the_ocean_girl help please

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