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katiebugg Group Title

ok simplify

  • 2 years ago
  • 2 years ago

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  1. katiebugg Group Title
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    \[a ^{2}\sqrt{8} /\sqrt{16a^{2}}\]

    • 2 years ago
  2. dumbcow Group Title
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    \[\sqrt{8} = \sqrt{4}*\sqrt{2}\] \[\sqrt{16a^{2}} = \sqrt{16}*\sqrt{a^{2}}\] that should help

    • 2 years ago
  3. katiebugg Group Title
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    not really lol sorry

    • 2 years ago
  4. dumbcow Group Title
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    what is sqrt(4) ?

    • 2 years ago
  5. katiebugg Group Title
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    what??

    • 2 years ago
  6. dumbcow Group Title
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    do you know what the the square root of 4 is?

    • 2 years ago
  7. dumbcow Group Title
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    ok how about 16?

    • 2 years ago
  8. katiebugg Group Title
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    2 and 4...

    • 2 years ago
  9. dumbcow Group Title
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    yep, now a^2

    • 2 years ago
  10. katiebugg Group Title
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    @dumbcow u lost me a long time ago

    • 2 years ago
  11. dumbcow Group Title
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    \[\sqrt{16} = \sqrt{4^{2}} = 4\] so for any number \[\sqrt{a^{2}} = a\]

    • 2 years ago
  12. katiebugg Group Title
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    lol ur not hepling my question tho...

    • 2 years ago
  13. dumbcow Group Title
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    well to simplify the problem , you need to simplify or get rid of the radicals, that is what im trying to help you do

    • 2 years ago
  14. katiebugg Group Title
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    dont u just multiply the top by the bottom???

    • 2 years ago
  15. panlac01 Group Title
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    you want someone to answer your problem? find the square root of 14 of a^2 and sqrt(4)*sqrt(2)

    • 2 years ago
  16. dumbcow Group Title
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    \[\frac{\sqrt{8} a^{2}}{\sqrt{16a^{2}}} = \frac{\sqrt{4}\sqrt{2}a^{2}}{\sqrt{16}\sqrt{a^{2}}} = \frac{2\sqrt{2}a^{2}}{4a}\]

    • 2 years ago
  17. katiebugg Group Title
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    @dumbcow ok now i seee ima sorry about that

    • 2 years ago
  18. dumbcow Group Title
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    np :) now just reduce to get final answer

    • 2 years ago
  19. katiebugg Group Title
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    lol thanks!!

    • 2 years ago
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