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anonymous

  • one year ago

Solve (z + 6)2 = 5 A. {-6+√5,6-√5} B. {-6±√5} C. {√5±-6} D. {6±√5}

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  1. anonymous
    • one year ago
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    is the 2 supposed to be a squared?

  2. anonymous
    • one year ago
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    Yes

  3. anonymous
    • one year ago
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    Okay. Thanks

  4. Australopithecus
    • one year ago
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    Rule you can multiple exponents on both sides of an equation so to solve for a: a^2 = b \[a^{2*(1/2)} = b^{1*(1/2)}\] \[a^{1} = b^{1/2}\] \[a = b^{1/2}\]

  5. Australopithecus
    • one year ago
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    It might also help to remember BEDMAS order of operations.

  6. anonymous
    • one year ago
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    \[\sqrt{(Z+6)^{2}}=\sqrt{5}\] Then solve for Z.

  7. anonymous
    • one year ago
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    Would it be D?

  8. Australopithecus
    • one year ago
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    I think you made a mistake in your algebra

  9. anonymous
    • one year ago
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    No, because if you subtract 6 from both sides the square root of 5 is not what is subtracted you subtract the 6.

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