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*_Artist_*

  • 3 years ago

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

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  1. ashish4deo
    • 3 years ago
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    64

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

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

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

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

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

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

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

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

  10. nowhereman
    • 3 years ago
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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.

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

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

  13. nowhereman
    • 3 years ago
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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.

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

  15. nowhereman
    • 3 years ago
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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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