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*_Artist_*

  • 5 years ago

True or false: 6 x2 - 2 x - 14 y + 3 x 1. There are 4 terms. 2. 6 x3 and 3 x are like terms. 3. The coefficient on y is 14. 4. Simplified, the expression is 6 x2 + 5 x - 14 y . 5. The commutative property allows the expression to be written as 6 x2 - 14 y + 3 x - 2 x . please help!

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  1. *_Artist_*
    • 5 years ago
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    oh! I forgot the was a exponent on one!

  2. *_Artist_*
    • 5 years ago
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    Sorry!

  3. nowhereman
    • 5 years ago
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    1. False there is only one term with a lot more subterms 2. What does "like terms" mean? 3. True 4. False the part with x should have coefficient 1 5. True

  4. *_Artist_*
    • 5 years ago
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    Thanks! Go to: http://wiki.answers.com/Q/What_is_a_like_term_in_math It gives a basic explanation.... Thank again!

  5. nowhereman
    • 5 years ago
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    That doesn't sound like math at all. Thus I can not give you the answer for 2. I mean on the one hand I could combine it like \[6x^3+3x = 3x(2x^2 + 1)\] but on the other hand that result still contains an adittion operation.

  6. *_Artist_*
    • 5 years ago
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    yeah...I realized the description isn't the best...I just Googled it... :\ It's okay...I think that they aren't....I'm not positive though....Thanks anyway! I fanned you!

  7. *_Artist_*
    • 5 years ago
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    BTW I messed up on the question...\[6x ^{2}\] is what it is supposed to be...

  8. radar
    • 5 years ago
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    1. False 2 Terms containing the same variable to the same power that can be combined like 4x, 7x , and 99x are like terms, x^2 is not the same as x and can't be directly combined. 3. No, the coefficient of y is -14 Note the negative sign. 4. If the 6 is intended to be a coefficient, yes. Why is the 6 set apart from the x^2??? 5. Yes What area do you need help in those examples?

  9. *_Artist_*
    • 5 years ago
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    Thanks! To answer your question from: question four, "If the 6 is intended to be a coefficient, yes. Why is the 6 set apart from the x^2???" It was: six times X with the exponent of 2...I'm not sure what you meant.... Thank you for helping me though!

  10. radar
    • 5 years ago
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    You're welcome anytime

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