anonymous
  • anonymous
c+66/c^2/1+4/c
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Jhannybean
  • Jhannybean
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.
anonymous
  • anonymous
|dw:1444169246249:dw|
Jhannybean
  • Jhannybean
Thank you :)

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Jhannybean
  • Jhannybean
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 \(\dfrac{c}{1}\)
Jhannybean
  • Jhannybean
We get : \[\frac{\dfrac{c^2}{c^2} \cdot c+\dfrac{64}{c^2}}{1+\dfrac{4}{c}}\]
Jhannybean
  • Jhannybean
The numerator will then simplify to \[\frac{\dfrac{c^3 +64}{c^2}}{1+\dfrac{4}{c}}\] Are you with me so far?
anonymous
  • anonymous
yes!
Jhannybean
  • Jhannybean
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: \(\dfrac{c+4}{c}\)
Jhannybean
  • Jhannybean
We're going to take \(\dfrac{c+4}{c}\) and replace \(1+\dfrac{4}{c}\) with this simplification :) \[\frac{\dfrac{c^3+64}{c^2}}{\dfrac{c+4}{c}}\]
anonymous
  • anonymous
gotcha
Jhannybean
  • Jhannybean
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: \(\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} \iff \dfrac{a}{b} \cdot \dfrac{d}{c} \)
Jhannybean
  • Jhannybean
Therefore: \[\frac{c^3+64}{c^2} \cdot \frac{c}{c+4}\]
Jhannybean
  • Jhannybean
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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