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baby456

  • one year ago

hey please help foe easy medal +fan square root of 9/16 +5

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  1. anonymous
    • one year ago
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    ??

  2. anonymous
    • one year ago
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    \[\sqrt{\frac{9}{16}} +5\] like that?

  3. baby456
    • one year ago
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    yes

  4. anonymous
    • one year ago
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    what is \(\sqrt{9}\)?

  5. baby456
    • one year ago
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    3

  6. anonymous
    • one year ago
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    right !

  7. anonymous
    • one year ago
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    and what is \(\sqrt{16}\)?

  8. triciaal
    • one year ago
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    |dw:1435706759551:dw|

  9. baby456
    • one year ago
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    is it rounded to 3/4

  10. anonymous
    • one year ago
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    no dear it is not "rounded" it is exactly \(\frac{3}{4}\)

  11. triciaal
    • one year ago
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    @satellite73 not your fan use both values

  12. anonymous
    • one year ago
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    \[\huge \sqrt{\frac{9}{16}}=\frac{\sqrt{9}}{\sqrt{16}}=\frac{3}{4}\]

  13. baby456
    • one year ago
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    ok thanks

  14. anonymous
    • one year ago
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    then add 5 i guess

  15. anonymous
    • one year ago
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    what the monkey does "use both values" mean?

  16. triciaal
    • one year ago
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    when you take the square root -3/4 and +3/4

  17. triciaal
    • one year ago
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    negative * negative = positive and positive * positive = positive

  18. anonymous
    • one year ago
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    ???

  19. myininaya
    • one year ago
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    sqrt(9/16) is the principal square root of 9/16 it is only the positive (or 0 if we have sqrt(0)) square root

  20. triciaal
    • one year ago
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    |dw:1435707126496:dw|

  21. anonymous
    • one year ago
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    really? i guess all this time the square root is not a well defined function. who knew?

  22. myininaya
    • one year ago
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    \[\sqrt{4}=2 \\ \text{ and if you had } \\ -\sqrt{4}=-(2)=-2\] but \[\sqrt{4} \neq -2\]

  23. anonymous
    • one year ago
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    elementary logic here \[x^2=4\implies x=2\text{ or } x=-2\] but \(\sqrt{4}\) is not some number whose square is 4, it is the positive number whose square is 4 you might have seen, for example, the \(\pm\) in the quadratic formula

  24. myininaya
    • one year ago
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    let me resay this: the square root of 4 is 2 or -2 but \[\sqrt{4}=2 \text{ and } \neq -2 \\ \text{ this is because we are suppose to read } \sqrt{ \cdot } \text{ as the princiapl square root }\]

  25. anonymous
    • one year ago
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    if you have a choice, you would not need to write \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] you could omit the \(\pm\) since apparently \(\sqrt{b^2-4ac}\) can be positive or negative

  26. Loser66
    • one year ago
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    Like this sentence "what the monkey does "use both values" mean?" hahahaha............

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