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anonymous

  • one year ago

Given the expression 3x2y + 4xy + 1 – 5x4y – 6y, do the following as instructed below: Write the polynomial in descending order. Classify the polynomial by the number of terms. State the degree of the polynomial.

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  1. DanJS
    • one year ago
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    every variable is to the first power?

  2. anonymous
    • one year ago
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    i dont get it

  3. DanJS
    • one year ago
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    you can combine things with the same variable terms.. like 3x2y + 4xy - 5x4y = 6xy + 4xy - 20 xy = (6 + 4 - 20)xy = -10xy

  4. anonymous
    • one year ago
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    does it apply for a.. b. nd c

  5. DanJS
    • one year ago
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    yes

  6. DanJS
    • one year ago
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    need to simplify it first by combining the like terms

  7. DanJS
    • one year ago
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    oh you know what, are those powers? like the first term \[3x^2*y\] ?

  8. anonymous
    • one year ago
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    NOPE . i dont know this at allllll

  9. DanJS
    • one year ago
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    They probably are, usually put a x^2 hat for powers

  10. DanJS
    • one year ago
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    So the prob is... \[3x^2y+4xy + 1 -5x^4 y - 6y\]

  11. DanJS
    • one year ago
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    Descending order - Largest powers of x first, Order the terms in decreasing powers of x

  12. DanJS
    • one year ago
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    Number of terms, are how many things are added or subtracted together

  13. DanJS
    • one year ago
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    the degree is the highest power

  14. anonymous
    • one year ago
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    so 5x4y is the degree

  15. anonymous
    • one year ago
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    what does it mean by classify for B

  16. DanJS
    • one year ago
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    sorry, the degree for terms like x^4*y is the sum of both variables exponents, here 4 + 1 = 5, the degree is 5

  17. DanJS
    • one year ago
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    just means how many terms are in the polynomial, here 5

  18. anonymous
    • one year ago
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    what is the polynomial by the number of terms

  19. DanJS
    • one year ago
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    each term is added or subtracted together

  20. DanJS
    • one year ago
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    like 3x^2*y is one term

  21. princesssleelee
    • one year ago
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    @DanJS can you explain this to me please? i have the same question

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