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gabriell Group Title

√(512)k^(2)

  • one year ago
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  1. ashlynn143 Group Title
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    i dont know

    • one year ago
  2. abb0t Group Title
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    \[\sqrt{a^1} = a^\frac{ 1 }{ 2 }\] now, in your example, you have: \[\sqrt{a^2} = a^\frac{ 2 }{ 2 } \] and you know: \[\frac{ 2 }{ 2 } = 1\]. And when you have \[\sqrt{4} = \sqrt{ 2 \times 2} = 2\] Similarly: \[\sqrt{100} = \sqrt{20 \times 5 } = \sqrt{(5 \times 4) \times 5} = \sqrt{(5 \times (2 \times 2)) \times 5} = \sqrt{(5 \times 5) (2 \times 2)}\] hence: \[\sqrt{(5 \times 5) (2 \times 2)} = 5 \times 2 = 10\]

    • one year ago
  3. abb0t Group Title
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    If you notice the patter, I am taking each pair of 5 and pair of 2 out of the radical. But only took out ONE PAIR of 5's and ONE PAIR of 2's. If you have more than one pair of 2's, then you would take out a 4. Because you know 2 x 2 = 4. and if you have two PAIRS of 5's you would take out a 25. Since you should know that 5 x 5 = 25.

    • one year ago
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