G^5 -g = g(g^4 -1)=g(g^2 -1)(g^2 +1) = g(g-1)(g+1)(g^2 +1)
24g^2 -6g^4 = 6g^2(4 -g^2) = 6g^2(2 -g)(2 +g)
A.
![5x-4=-2(3x+2) \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\ 5x-4=-2 \times 3x-2 \times 2 \\ 5x-4=-6x-4 \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 6x to both sides} \\ 11x-4=-4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 4 to both sides} \\ 11x=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{divide both sides by 11} \\ x=0](https://tex.z-dn.net/?f=5x-4%3D-2%283x%2B2%29%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%7C%5Chbox%7Bexpand%20the%20bracket%7D%20%5C%5C%0A5x-4%3D-2%20%5Ctimes%203x-2%20%5Ctimes%202%20%5C%5C%0A5x-4%3D-6x-4%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%7C%5Chbox%7Badd%206x%20to%20both%20sides%7D%20%5C%5C%0A11x-4%3D-4%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%7C%5Chbox%7Badd%204%20to%20both%20sides%7D%20%5C%5C%0A11x%3D0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%7C%5Chbox%7Bdivide%20both%20sides%20by%2011%7D%20%5C%5C%0Ax%3D0)
B.
![3(2x-4)=5x-1 \\ 6x-12=5x-1 \\ \boxed{11x-12=-1} \Leftarrow \hbox{the first mistake} \\ 11x=11 \\ \boxed{x=11} \Leftarrow \hbox{the second mistake}](https://tex.z-dn.net/?f=3%282x-4%29%3D5x-1%20%5C%5C%0A6x-12%3D5x-1%20%5C%5C%0A%5Cboxed%7B11x-12%3D-1%7D%20%5CLeftarrow%20%5Chbox%7Bthe%20first%20mistake%7D%20%5C%5C%0A11x%3D11%20%5C%5C%0A%5Cboxed%7Bx%3D11%7D%20%5CLeftarrow%20%5Chbox%7Bthe%20second%20mistake%7D)
Megan's solution isn't correct.
The first mistake: she subtracted 5x from the right-hand side of the equation, but added 5x to the left-hand side.
The second mistake: she divided the right-hand side of the equation by 11, but didn't divide the left-hand side.
The correct solution:
The significance level, also denoted as alpha or α, is the probability of rejecting the null hypothesis when it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference.
A. 8/$12=21/$x then cross multiply.
B. 1/$1.5