$\displaystyle 1/3x^2 - 5/6x =1 $
$\displaystyle x^3 -26x ^3 / ^2 =27 $
$\displaystyle 2x^2(2x-1) ^1/ ^3 =4x(2x-1) ^4 / ^3 $
$\displaystyle 15/ 2^x -1 = 7 $ this reads as 15 ALL OVER 2^x -1
$\displaystyle (1/3)x^2-(5/6)x=1 $
$\displaystyle (x)^3-(26)(x ^ 3 / ^ 2) = 27 $
$\displaystyle (2x^2)(2x-1)(^1/^3) = (4x)(2x-1) (^4/ ^3) $
$\displaystyle (15) / (2^x - 1) =7 $
i hopw this is better ... sorry about that
AP Calculus Ab 2011 Summer Work Part 1
if you're having trouble still reading it please open this document and look at #'s 2, 10,11,15,37, and 40 ( the same exact problems i need help)
any help on this problems would be appreciated!!