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anonymous

  • one year ago

c+66/c^2/1+4/c

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  1. Jhannybean
    • one year ago
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    Do you mind drawing out your question? Im not understanding which ones are fractions and which arent. Your lack of parenthesis makes it hard to differentiate that.

  2. anonymous
    • one year ago
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    |dw:1444169246249:dw|

  3. Jhannybean
    • one year ago
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    Thank you :)

  4. Jhannybean
    • one year ago
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    First we need to find a common denominator in both the numerator and denominator. For the numerator, the LCD between \(1\) and \(c^2\) is \(c^2\). Let's multiply \(c^2\) to the top and bottom of \(\dfrac{c}{1}\)

  5. Jhannybean
    • one year ago
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    We get : \[\frac{\dfrac{c^2}{c^2} \cdot c+\dfrac{64}{c^2}}{1+\dfrac{4}{c}}\]

  6. Jhannybean
    • one year ago
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    The numerator will then simplify to \[\frac{\dfrac{c^3 +64}{c^2}}{1+\dfrac{4}{c}}\] Are you with me so far?

  7. anonymous
    • one year ago
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    yes!

  8. Jhannybean
    • one year ago
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    Great. Now let's find the LCD in the denominator. The LCD in the denominator between \(1\) and \(c\) will be \(c\), therefore we multiply \(c\) to both the numerator and denominator of 1. \[\frac{c}{c} \cdot 1 + \frac{4}{c}\] This becomes: \(\dfrac{c+4}{c}\)

  9. Jhannybean
    • one year ago
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    We're going to take \(\dfrac{c+4}{c}\) and replace \(1+\dfrac{4}{c}\) with this simplification :) \[\frac{\dfrac{c^3+64}{c^2}}{\dfrac{c+4}{c}}\]

  10. anonymous
    • one year ago
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    gotcha

  11. Jhannybean
    • one year ago
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    When we're dividing 2 fractions, we're going to multiply by the reciprocal of the second fraction to the first fraction. It follows this simple rule: \(\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} \iff \dfrac{a}{b} \cdot \dfrac{d}{c} \)

  12. Jhannybean
    • one year ago
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    Therefore: \[\frac{c^3+64}{c^2} \cdot \frac{c}{c+4}\]

  13. Jhannybean
    • one year ago
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    We're going to cross cancel our like terms but if we take a look, \(c^3+64\) can be expanded since it is of the form \(a^3+b^3\). Do you follow?

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