B^2*b^3 over (2b^2)^2

There was nothing to "solve"......remember........we have to have an " = "  sign in the problem to do that......I just  "simplified"  it using some basic rules involving exponents.....

well 2 bits. I didnt know that was actually possible to calculate. Sounds like a networking problem? Maybe just the bits part throws me off.

 [B^2*b^3]  / (2b^2)^2  =

[ B^2*b^3 ] / [4 b^4]  =

[B^2 ]  / [ 4b ]

hey cphill what was that math you just did called?

Just Basic Algebra, Z-Man........

B^2*b^3 over (2b^2)^2 we never used B as a unit. What does it stand for?

I don't know, zegroes......it's just a variable.......and, since it's upper-case, we usually assume that it's different from the lower-case "b"......What  is  each suppposed to represent?????.......who knows????

So how did you solve it without knowing what the variables were?

alright i get it. Thanks fossil. Who knew i could learn something from an old person....How the times have changed

ROF LOL