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marixoxo76

  • 3 years ago

13r^-4(r^-5)^5/8(r^2)^-2 solve with steps please.

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  1. marixoxo76
    • 3 years ago
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    \[\frac{ 13r ^{-4}(r ^{-5})^{5} }{ 8(r ^{2})^{-2} }\]

  2. nickersia
    • 3 years ago
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    \[\frac{ 13*\frac{ 1 }{ r ^{4} } (\frac{ 1 }{ r ^{5} })^{5} }{ 8*\frac{ 1 }{ (r ^{2})^{2} } }\] Do you know how to do that?

  3. marixoxo76
    • 3 years ago
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    Not really... can you explain please?

  4. nickersia
    • 3 years ago
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    \[r ^{-100} = \frac{ 1 }{ r ^{100} }\] Is that clear?

  5. nickersia
    • 3 years ago
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    \[r ^{-4} = \frac{ 1 }{ r ^{4} }\]

  6. nickersia
    • 3 years ago
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    So now you have to simplify it \[\frac{ 13 *\frac{ 1 }{ r ^{4} * r ^{25} } }{ 8 * \frac{ 1 }{ r ^{4} } }\] 25 because 5*5 = 25 \[\frac{ 13*\frac{ 1 }{ r ^{29} } }{ 8 * \frac{ 1 }{ r ^{4} } }\] \[\frac{ \frac{ 13 }{ r ^{29} } }{ \frac{ 8 }{ r ^{4} } }\] (very up number is 13, hard to read) \[\frac{ 13 * r ^{4} }{ 8 * r ^{29} }\] 29-4=25 \[\frac{ 13 }{ 8r ^{25} }\] That should be resoult :)

  7. marixoxo76
    • 3 years ago
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    ah, perfect! thank you so much i was struggling with problems like these thanks got it :)

  8. nickersia
    • 3 years ago
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    Nice to hear that! Good luck!

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