Thread: Please help!! solve for x

1. Please help!! solve for x

$\displaystyle 1/3x^2 - 5/6x =1$

$\displaystyle x^3 -26x ^3 / ^2 =27$

$\displaystyle 2x^2(2x-1) ^1/ ^3 =4x(2x-1) ^4 / ^3$

$\displaystyle 15/ 2^x -1 = 7$ this reads as 15 ALL OVER 2^x -1

2. Re: Please help!! solve for x

Do you not have parentheses keys on your keyboard? You posted similar questions (in several cases exactly same equations) yesterday and were told why it was impossible to be certain what you meant.

3. Re: Please help!! solve for x

$\displaystyle (1/3)x^2-(5/6)x=1$

$\displaystyle (x)^3-(26)(x ^ 3 / ^ 2) = 27$

$\displaystyle (2x^2)(2x-1)(^1/^3) = (4x)(2x-1) (^4/ ^3)$

$\displaystyle (15) / (2^x - 1) =7$

i hopw this is better ... sorry about that

4. Re: Please help!! solve for x

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!!

5. Re: Please help!! solve for x

looks like a graded assignment ... thread closed.