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anonymous
 5 years ago
Solve for V in V = s^3 , if s = 4
anonymous
 5 years ago
Solve for V in V = s^3 , if s = 4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was wondering how he got that...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A= 1/2 h \[ ( b _{1} + b _{2} )\] if A = 16, h = 4, and b1 = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so basically 16 = 1/2*4 (3+\[ b _{2}\])

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know about equivalent transformation of equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Uh...I'm not sure....could you please explain for me?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0If you have a function which is injective (i.e. onetoone) 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 (nonzero) factor on both sides, because you can invert that operation by dividing through that factor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How would I use it on my problem?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what would be the inverse for \[b _{2}\]?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. On the first side we have 8 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Square root is the inverse of the power of two.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0There is no power of two anywhere here^^ yes, 8 is correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. How do I divide the second half by 2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, I wrongly assumed \[b_2\] was a typo of \[b^{2}\] Sorry. :)

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0First multiplying something (3+b2 e.g.) by 2 and then dividing by two again, what will you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the first answer....?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what you are referring to...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"First multiplying something (3+b2 e.g.) by 2 and then dividing by two again, what will you get?"

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0You just have to think about that something as a number.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, now what am i supposed to do? I'm confused.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Try to answer that last question, if you don't know it, try it out with several numbers.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i gotta go.... be back later
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