1. #36 is wrong: You used the wrong equation
2. #37 is OK now
3. #38 is wrong. As I mentioned in my previous post you have to determine the appropriate domain. From
$\displaystyle x^3+|y|=6~\implies~|y|=6-x^3$
Since $\displaystyle |y|\geq 0$ you have to calculate those values for x that
$\displaystyle 6-x^3\geq 0~\implies~x^3 \leq 6~\implies~\boxed{x\leq \sqrt[3]{6}}$
That means $\displaystyle |y|=6-x^3\ ,\ x\leq \sqrt[3]{6}$ is a function.
Definitely I won't do your homework. So show some work of your own and if there are some mistakes you'll get additional support....also i want solve questions from 39 to 42
What you labeled #36 is actually #37. Usually an equation of a function starts with "y = ...". So your result is slightly unfinished (but correct)
#39 is wrong.
Explanation: $\displaystyle \left(\dfrac0a = 0\ ,\ a \in \mathbb{R} \setminus \{0\}\right)$
#40 is OK
#41 is wrong. You only have to solve the given equation for y to judge wether the equation describes a function or not.
#42 is correct but incomplete. (Is this a function or not?)
Initially you are asked to find the domain of those equations which define a function. The domains are missing
you mean in #39 the equation of a function withou ( - ) or what ?
what is wrong here ?
#39
$\displaystyle
\left(\dfrac0X = 0\ ,\ X \in \mathbb{R} \setminus \{0\}\right)
$ function
is ok now ?
#41
x + y = 1
y = 1- x
function
#42 a function