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anonymous

  • one year ago

Find the value of the following expression

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @danjs

  3. anonymous
    • one year ago
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    @Nnesha @jhannybean @mathstudent55

  4. DanJS
    • one year ago
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    have to use a couple of the exponent rules

  5. DanJS
    • one year ago
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    \[\large [u^a]^b = u^{a*b}\] \[u^a * u^b = u^{a+b}\]

  6. anonymous
    • one year ago
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    @danjs I know that but I cant seem to get the answer right

  7. DanJS
    • one year ago
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    k, let me enter the thing

  8. DanJS
    • one year ago
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    \[\large (2^8 * 3^{-5}*6^0)^{-2}*[\frac{ 3^{-2} }{ 2^3 }]^4*2^{28}\]

  9. anonymous
    • one year ago
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    @DanJS I know what the expression is I need the answer...

  10. DanJS
    • one year ago
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    i can give you the answer, but how would you know where you went wrong to get there

  11. anonymous
    • one year ago
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    ill ask my teacher tom

  12. DanJS
    • one year ago
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    i can show you how to apply the exponent properties and get to the answer

  13. DanJS
    • one year ago
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    You can only apply the exponent rules if the base is the same... I would do the parenthesis part first, and a power raised to a power is where you multiply the powers to simplify \[\large (2^8 * 3^{-5}*6^0)^{-2}*\frac{ 3^{-2*4} }{ 2^{3*4} }*2^{28}\]

  14. anonymous
    • one year ago
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    -40?

  15. DanJS
    • one year ago
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    The parenthesis part left has 6^0 in it, something to the zero power is always 1. Here again you apply the power raised to a power rule to simplify \[\large (2^{-16} * 3^{10}*1)*\frac{ 3^{-8} }{ 2^{12} }*2^{28}\]

  16. DanJS
    • one year ago
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    not sure, i didnt calculate anything

  17. DanJS
    • one year ago
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    You can change the side of a fraction an exponential is on by changing the sign of it's exponent, move the 2^12 to the numorator, you are left with just a string of terms being multiplied

  18. DanJS
    • one year ago
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    \[\large 2^{-16} * 3^{10}*1* 3^{-8}* 2^{-12} *2^{28}\] combine the powers of the like bases, then it is simplified

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