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anonymous

  • one year ago

24. Simplify the complex fraction. (5x^3/x+2)/30x^2/x+3) A) x(x+3)/6(x+2) B) 6x/ (x+2)(x+3) C) 150x^5/(x+2)(x+3) D) x+3/2 25. Divide the following rational expression, 16z^5 - 12z^4 + 8z^3 - 6z/2z^2 A) 8z^3 - 6z^2 + 8z - 3z B) 8z^3 - 6z^2 + 4z - 3/z C) 8z^3 + 6z^2 + 8z + 3z D) 10z^5 - 3

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  1. Vocaloid
    • one year ago
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    @Directrix I'm about to get some sleep, would you mind taking over?

  2. Jhannybean
    • one year ago
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    \[\large \frac{\dfrac{5x^3}{x+2}}{\dfrac{30x^2}{x+3}}\] Remember, when you're dividing by 2 fractions, follow this rule: \[\large \frac{\frac{a}{b}}{\frac{c}{d}} \implies \frac{a}{b} \cdot \frac{d}{c}\]

  3. Jhannybean
    • one year ago
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    therefore, \( \frac{\dfrac{5x^3}{x+2}}{\dfrac{30x^2}{x+3}}\) becomes...\[=\frac{5x^3}{x+2} \cdot \frac{x+3}{30x^2}\]

  4. Jhannybean
    • one year ago
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    Now we just cross cancel like terms, \[\frac{\color{red}{5}\color{blue}{x^3}}{x+2} \cdot \frac{x+3}{\color{red}{30}\color{blue}{x^2}}\]

  5. Jhannybean
    • one year ago
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    For your second problem: \(\dfrac{16z^5 - 12z^4 + 8z^3 - 6z}{2z^2}\) Divide each term in the numerator by the term in the denominator. \[\frac{\color{red}{16}\color{blue}{z^5}}{\color{red}{2}\color{blue}{z^2}} -\frac{\color{red}{12}\color{blue}{z^4}}{\color{reD}{2}\color{blue}{z^2}}+\frac{\color{red}{8}\color{blue}{z^3}}{\color{red}{2}\color{blue}{z^2}}-\frac{\color{red}{6}\color{blue}{z}}{\color{red}{2}\color{blue}{z^2}}\]

  6. anonymous
    • one year ago
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    @Jhannybean Thanks, I've completed them. :)

  7. Jhannybean
    • one year ago
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    Np :)

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