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Solve for V in V = s^3 , if s = 4
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ashish4deo
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64
knowak
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s^3 = s * s * s
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Thank you guys.
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I was wondering how he got that...
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A= 1/2 h \[ ( b _{1} + b _{2} )\] if A = 16, h = 4, and b1 = 3
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so basically 16 = 1/2*4 (3+\[ b _{2}\])
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I need to find b2
nowhereman
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Do you know about equivalent transformation of equations?
*_Artist_*
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Uh...I'm not sure....could you please explain for me?
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.
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okay..
*_Artist_*
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How would I use it on my problem?
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.
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what would be the inverse for \[b _{2}\]?
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.
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Okay. On the first side we have 8 right?
knowak
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Square root is the inverse of the power of two.
nowhereman
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There is no power of two anywhere here^^
yes, 8 is correct.
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Okay. How do I divide the second half by 2?
knowak
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Ah, I wrongly assumed \[b_2\] was a typo of \[b^{2}\] Sorry. :)
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s'okay! ^^
nowhereman
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First multiplying something (3+b2 e.g.) by 2 and then dividing by two again, what will you get?
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the first answer....?
nowhereman
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I'm not sure what you are referring to...
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i'm confused now too
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"First multiplying something (3+b2 e.g.) by 2 and then dividing by two again, what will you get?"
nowhereman
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You just have to think about that something as a number.
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okay...
*_Artist_*
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so, now what am i supposed to do? I'm confused.
nowhereman
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Try to answer that last question, if you don't know it, try it out with several numbers.
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which one?
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i gotta go....
be back later