SOLUTION: I understand that i have to multiply so that the denomomintor is 1 but I dont know how to do that because the denominator is a binomal. Heres the problem, {{{b^-.5}}} over {{{

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I understand that i have to multiply so that the denomomintor is 1 but I dont know how to do that because the denominator is a binomal. Heres the problem, {{{b^-.5}}} over {{{      Log On


   

Question 59877This question is from textbook Algebra 2
: I understand that i have to multiply so that the denomomintor is 1 but I dont know how to do that because the denominator is a binomal.
Heres the problem,
b%5E-.5 over
8b%5E.333b%5E-.25
If you have the time it would be appreciated if you could sohw me the steps This question is from textbook Algebra 2

Found 2 solutions by funmath, stanbon:
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
b%5E%28-.5%29%2F%288b%5E.333%2Ab%5E%28-.25%29%29
The first thing I would do is move the things with negative exponents where they aren't (either numerator or denominator) and change the decimals to fractions.
b%5E%28.25%29%2F%288b%5E.333%2Ab%5E%28.5%29%29
b%5E%281%2F4%29%2F%288b%5E%281%2F3%29%2Ab%5E%281%2F2%29%29
b%5E%281%2F4%29%2F%288b%5E%281%2F3%2B1%2F2%29%29
b%5E%281%2F4%29%2F%288b%5E%282%2F6%2B3%2F6%29%29
b%5E%281%2F4%29%2F%288b%5E%285%2F6%29%29
%28b%5E%281%2F4%29%2Ab%5E%281%2F6%29%29%2F%288b%5E%285%2F6%29%2Ab%5E%281%2F6%29%29
b%5E%281%2F4%2B1%2F6%29%2F%288b%5E%285%2F6%2B1%2F6%29%29
b%5E%283%2F12%2B2%2F12%29%2F%288b%5E%286%2F6%29%29
b%5E%285%2F12%29%2F8b
:
If your problem was really in decimal form to begin with:
b%5E%28-.5%29%2F%288b%5E.333%2Ab%5E%28-.25%29%29
b%5E%28.25%29%2F%288b%5E.333%2Ab%5E.5%29
b%5E.25%2F%288b%5E%28.333%2B.5%29%29
b%5E.25%2F%288b%5E%28.833%29%29
b%5E.25%2Ab%5E.167%2F%288b%5E.833%2Ab%5E.167%29
b%5E%28.25%2B.167%29%2F8b%5E1
b%5E.417%2F8b
I am assuming that the denominator can't have a decimal in its exponent and it can't have a negative exponent.
Happy Calculating!!!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
b^-.5 over 8b^.333b^-.25
=(1/8)b^(-0.5-0.333+0.25)
=(1/8)b^-0.583
=1/[8b^0.583]
Cheers,
Stan H.