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BelleFlower

  • 3 years ago

Please refer to image! Indices question

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  1. hba
    • 3 years ago
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    k

  2. BelleFlower
    • 3 years ago
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    Help with Q12! :)

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  3. mark_o.
    • 3 years ago
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    \[P=(2+\sqrt{5})^{2n-1}\] \[P=\frac{ (2+\sqrt{5})^{2n}(2-\sqrt{5}) }{ (2+\sqrt{5})(2-\sqrt{5}) }\] \[P=\frac{ [(2+\sqrt{5})^{2}]^{n}(2-\sqrt{5}) }{ 4-5 }\]

  4. mark_o.
    • 3 years ago
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    can you try now the next step?

  5. mark_o.
    • 3 years ago
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    @BelleFlower

  6. mark_o.
    • 3 years ago
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    \[P=\frac{ (9+4\sqrt{5})^{n}(2+\sqrt{5}) }{ -1 }\] P=\[P=(\sqrt{5}-2)(9+4\sqrt{5})^{n}\]

  7. mark_o.
    • 3 years ago
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    you can continue the next step

  8. mark_o.
    • 3 years ago
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    thats ok :D

  9. mark_o.
    • 3 years ago
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    check on this one \[\sqrt{(4)^{2}}\sqrt{5}=\sqrt{16*5}=\sqrt{80}\]

  10. BelleFlower
    • 3 years ago
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    OH OH okay thank you so much! I got it now!!

  11. mark_o.
    • 3 years ago
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    YW good luck now ... :D

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