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

- anonymous

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

- schrodinger

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- anonymous

64

- anonymous

s^3 = s * s * s

- anonymous

Thank you guys.

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## More answers

- anonymous

I was wondering how he got that...

- anonymous

A= 1/2 h \[ ( b _{1} + b _{2} )\] if A = 16, h = 4, and b1 = 3

- anonymous

so basically 16 = 1/2*4 (3+\[ b _{2}\])

- anonymous

I need to find b2

- nowhereman

Do you know about equivalent transformation of equations?

- anonymous

Uh...I'm not sure....could you please explain for me?

- nowhereman

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.

- anonymous

okay..

- anonymous

How would I use it on my problem?

- nowhereman

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.

- anonymous

what would be the inverse for \[b _{2}\]?

- nowhereman

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.

- anonymous

Okay. On the first side we have 8 right?

- anonymous

Square root is the inverse of the power of two.

- nowhereman

There is no power of two anywhere here^^
yes, 8 is correct.

- anonymous

Okay. How do I divide the second half by 2?

- anonymous

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

- anonymous

s'okay! ^^

- nowhereman

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

- anonymous

the first answer....?

- nowhereman

I'm not sure what you are referring to...

- anonymous

i'm confused now too

- anonymous

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

- nowhereman

You just have to think about that something as a number.

- anonymous

okay...

- anonymous

so, now what am i supposed to do? I'm confused.

- nowhereman

Try to answer that last question, if you don't know it, try it out with several numbers.

- anonymous

which one?

- anonymous

i gotta go....
be back later

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