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anonymous

  • one year ago

Part A: Divide (8x4y3 + 4x3y2 - 2x2y - 12x2y4) by -2x2y. Show your work, and justify each step. Part B: How would your answer in Part A be affected if the x2 variable in the denominator was just an x? Part C: What is the degree and classification of the polynomial you got in Part A?

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  1. anonymous
    • one year ago
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    @phi please help

  2. anonymous
    • one year ago
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    someone

  3. phi
    • one year ago
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    you divide each "term" by the denominator can you do the first term \[ \frac{8x^4y^3 }{-2x^2y}\]

  4. phi
    • one year ago
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    first do the numbers 8/(-2) then x*x*x*x/x*x then y*y*y/y

  5. anonymous
    • one year ago
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    what?

  6. phi
    • one year ago
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    Divide (8x4y3 + 4x3y2 - 2x2y - 12x2y4) by -2x2y. that means you divide each term by -2x2y the first term is 8x4y3 divided by -2x2y

  7. phi
    • one year ago
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    as you know, \[ \frac{8x^4y^3 }{-2x^2y}= \frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y} \]

  8. anonymous
    • one year ago
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    ok so first is 8/-2 which equals -4

  9. phi
    • one year ago
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    or, if we expand the exponents for x \[ \frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y} \\ \frac{8}{-2} \cdot \frac{x\cdot x \cdot x \cdot x}{x\cdot x} \cdot \frac{y^3}{y} \]

  10. phi
    • one year ago
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    if you have the "same thing" in the "top" and "bottom" they divide out (give you 1) so every x/x can be cancelled.

  11. anonymous
    • one year ago
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    4*256*27

  12. anonymous
    • one year ago
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    is this right

  13. phi
    • one year ago
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    letters stay letters

  14. anonymous
    • one year ago
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    I have 10 mins and about 20 secs left can you try and help a little faster sorry for rushing I just don't understand

  15. phi
    • one year ago
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    we use the idea that anything divided by itself is 1 \[ \frac{x \cdot x}{x \cdot x} = 1 \]

  16. phi
    • one year ago
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    so \[ \frac{x^4}{x^2} = \frac{x\cdot x \cdot x\cdot x}{x\cdot x}= x^2 \]

  17. phi
    • one year ago
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    now do the y^3/y

  18. anonymous
    • one year ago
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    y3

  19. phi
    • one year ago
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    \[ \frac{y \cdot y \cdot y}{y} \] you have one y/y pair that you can cancel

  20. phi
    • one year ago
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    you should get \[ \frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y} \\ -4 x^2y^2\]

  21. phi
    • one year ago
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    now do the next term 4x3y2 divided by -2x2y

  22. anonymous
    • one year ago
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    -2 x^1.5 y^2

  23. anonymous
    • one year ago
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    @phi

  24. anonymous
    • one year ago
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    @phi is the answer -4^2y^2-2x^1.5y^26xy^4

  25. anonymous
    • one year ago
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    @DarronW do you know if this is the answer

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