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joyce153

  • 4 years ago

See attachment.

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  1. joyce153
    • 4 years ago
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  2. Hunus
    • 4 years ago
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    \[a^{12}*a^{0}=a^{12}*1=a^{12}\]

  3. meverett04
    • 4 years ago
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    What can we do with the 12 and 0? Hint: \[x^3 \times x^5 = x^8\]

  4. joyce153
    • 4 years ago
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    so the answer is \[a ^{12}\]

  5. Hunus
    • 4 years ago
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    Yes, because \[a^{m}*a^{n}=a^{m+n}\]\[a^{12}*a^{0}=a^{12+0}=a^{12}\]

  6. joyce153
    • 4 years ago
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    kk THX

  7. Hunus
    • 4 years ago
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    no problem :)

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