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*_Artist_*

Solve for V in V = s^3 , if s = 4

  • 2 years ago
  • 2 years ago

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  1. ashish4deo
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    64

    • 2 years ago
  2. knowak
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    s^3 = s * s * s

    • 2 years ago
  3. *_Artist_*
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    Thank you guys.

    • 2 years ago
  4. *_Artist_*
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    I was wondering how he got that...

    • 2 years ago
  5. *_Artist_*
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    A= 1/2 h \[ ( b _{1} + b _{2} )\] if A = 16, h = 4, and b1 = 3

    • 2 years ago
  6. *_Artist_*
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    so basically 16 = 1/2*4 (3+\[ b _{2}\])

    • 2 years ago
  7. *_Artist_*
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    I need to find b2

    • 2 years ago
  8. nowhereman
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    Do you know about equivalent transformation of equations?

    • 2 years ago
  9. *_Artist_*
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    Uh...I'm not sure....could you please explain for me?

    • 2 years ago
  10. nowhereman
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    If you have a function which is injective (i.e. one-to-one) you can apply it to both sides of an equation and get an equivalent expression. i.e. the new expression is true for exactly the same allocation of the variables with values as the initial one. So for example you can multiply with a constant (non-zero) factor on both sides, because you can invert that operation by dividing through that factor.

    • 2 years ago
  11. *_Artist_*
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    okay..

    • 2 years ago
  12. *_Artist_*
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    How would I use it on my problem?

    • 2 years ago
  13. nowhereman
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    You cant to seperate b2, but you have it enclosed in operations, so you must apply the inverses of those functions and as explained before, to get the correct b2 you have to do that on both sides.

    • 2 years ago
  14. *_Artist_*
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    what would be the inverse for \[b _{2}\]?

    • 2 years ago
  15. nowhereman
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    You have \[16 = 2(3 + b_2)\] given. So for a start the outermost operation on the right hand side is multiplication by 2. So first you would have to divide by 2 on both sides.

    • 2 years ago
  16. *_Artist_*
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    Okay. On the first side we have 8 right?

    • 2 years ago
  17. knowak
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    Square root is the inverse of the power of two.

    • 2 years ago
  18. nowhereman
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    There is no power of two anywhere here^^ yes, 8 is correct.

    • 2 years ago
  19. *_Artist_*
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    Okay. How do I divide the second half by 2?

    • 2 years ago
  20. knowak
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    Ah, I wrongly assumed \[b_2\] was a typo of \[b^{2}\] Sorry. :)

    • 2 years ago
  21. *_Artist_*
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    s'okay! ^^

    • 2 years ago
  22. nowhereman
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    First multiplying something (3+b2 e.g.) by 2 and then dividing by two again, what will you get?

    • 2 years ago
  23. *_Artist_*
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    the first answer....?

    • 2 years ago
  24. nowhereman
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    I'm not sure what you are referring to...

    • 2 years ago
  25. *_Artist_*
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    i'm confused now too

    • 2 years ago
  26. *_Artist_*
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    "First multiplying something (3+b2 e.g.) by 2 and then dividing by two again, what will you get?"

    • 2 years ago
  27. nowhereman
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    You just have to think about that something as a number.

    • 2 years ago
  28. *_Artist_*
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    okay...

    • 2 years ago
  29. *_Artist_*
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    so, now what am i supposed to do? I'm confused.

    • 2 years ago
  30. nowhereman
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    Try to answer that last question, if you don't know it, try it out with several numbers.

    • 2 years ago
  31. *_Artist_*
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    which one?

    • 2 years ago
  32. *_Artist_*
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    i gotta go.... be back later

    • 2 years ago
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