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ConnexussKidd96

  • 3 years ago

Im so lost Can someone please help me with this geomtric mean.. ive started the equation but how to you find the square root? Question BELOW

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  1. ConnexussKidd96
    • 3 years ago
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    \[\frac{ 5 }{ x }\frac{ x }{ 12} x^2=60 x=\sqrt{60}\]

  2. hartnn
    • 3 years ago
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    huh ? do u just need square root of 60 ?

  3. ConnexussKidd96
    • 3 years ago
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    yes how to you get the square root?

  4. hartnn
    • 3 years ago
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    can you factor 60 ?

  5. ConnexussKidd96
    • 3 years ago
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    what does that mean

  6. hartnn
    • 3 years ago
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    like 10=2*5 24=2*2*2*3

  7. hartnn
    • 3 years ago
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    60=...?

  8. ConnexussKidd96
    • 3 years ago
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    i dont ge it how did you get those numbers

  9. hartnn
    • 3 years ago
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    divide 60 by 2, what u get ?

  10. ConnexussKidd96
    • 3 years ago
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    30

  11. hartnn
    • 3 years ago
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    so, 60 = 2*30 now 30 is a prime number ? no. so again divide 30 by 2, what u get ?

  12. ConnexussKidd96
    • 3 years ago
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    15

  13. hartnn
    • 3 years ago
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    so, 60 = 2*2*15 now 15 is a prime number ? no. so again divide 15 by 2, can't divide.... divide 15 by 3, what u get ?

  14. ConnexussKidd96
    • 3 years ago
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    5

  15. hartnn
    • 3 years ago
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    so, 60 = 2*2*3*5 right ? (since, 5 is a prime number, we stop) are you getting all this ?

  16. ConnexussKidd96
    • 3 years ago
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    yes right !! & yes i am

  17. hartnn
    • 3 years ago
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    now comes the easy part ... \[\sqrt{x.x.y}=\sqrt{x.x}\sqrt y \] so what about \[\sqrt{60}=\sqrt{2.2.3.5}=...?\]

  18. hartnn
    • 3 years ago
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    i meant \[\sqrt{x.x.y}=\sqrt{x.x}\sqrt y=x\sqrt y\]

  19. ConnexussKidd96
    • 3 years ago
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    wow sorry im horrible at math but is it \[2\sqrt{15}\]

  20. hartnn
    • 3 years ago
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    yes, its absolutely correct :) and you are not horrible at math, you understood all this, means you are good :)

  21. hartnn
    • 3 years ago
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    any doubts ?

  22. ConnexussKidd96
    • 3 years ago
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    No Doubts!! & Thannk you so much for helping

  23. hartnn
    • 3 years ago
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    welcome ^_^

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