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anonymous

  • 4 years ago

find LCM...10b^4c^9, 2b^6c^8, 15b^7c^6

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  1. anonymous
    • 4 years ago
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    ok im confused on how to write the problem out. The way I did it was \[2\times5.b.b.b.b.b.b.b.b.b.b.b.b.b\]

  2. anonymous
    • 4 years ago
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    the way i did it was what two numbers equal 10 (2x5) then i thought we write out how many b's or c's there are

  3. Hero
    • 4 years ago
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    Forget what I wrote previously.

  4. Hero
    • 4 years ago
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    Instead, think of trying to find the Least Common Denominator

  5. anonymous
    • 4 years ago
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    i did that with each number

  6. Hero
    • 4 years ago
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    \[\frac{1}{10b^4c^9}+\frac{1}{2b^6c^8}+\frac{1}{15b^7c^6}\] Imagine you had to add those up. How would you find the LCD?

  7. anonymous
    • 4 years ago
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    would that be 2?

  8. Hero
    • 4 years ago
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    30b^7c^9 is the correct answer here.

  9. anonymous
    • 4 years ago
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    ok i have no clue how you got that lol

  10. anonymous
    • 4 years ago
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    so the way i was working it, isnt a good way?

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spraguer (Moderator)
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