1. I want check my answer and solve some question

I want check my answer and solve some question

Q :

my answer from 33 - 38

thanks

2. Originally Posted by r-soy
I want check my answer and solve some question

Q :

my answer from 33 - 38

thanks
You did correctly the questions #33, #34, #35

The last 3 questions are wrong:
#36: You used the wrong equation
#37 no root to be used
#38 use definition of absolute value and determine the appropriate domain

3. Originally Posted by earboth
You did correctly the questions #33, #34, #35

The last 3 questions are wrong:
#36: You used the wrong equation
#37 no root to be used
#38 use definition of absolute value and determine the appropriate domain
thanks ,,,

check it now

earboth also i want solve questions from 39 to 42

thanks >>

4. Originally Posted by r-soy
thanks ,,,

check it now

earboth also i want solve questions from 39 to 42

thanks >>
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

$x^3+|y|=6~\implies~|y|=6-x^3$

Since $|y|\geq 0$ you have to calculate those values for x that
$6-x^3\geq 0~\implies~x^3 \leq 6~\implies~\boxed{x\leq \sqrt[3]{6}}$

That means $|y|=6-x^3\ ,\ x\leq \sqrt[3]{6}$ is a function.

...also i want solve questions from 39 to 42
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.

5. Now

i try ::::

6. Originally Posted by r-soy
Now

i try ::::
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: $\left(\dfrac0a = 0\ ,\ a \in \mathbb{R} \setminus \{0\}\right)$

#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

7. you mean in #39 the equation of a function withou ( - ) or what ?
what is wrong here ?

#39
$
\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

8. Originally Posted by r-soy
you mean in #39 no I meant #36 the equation of a function withou ( - ) or what ?
what is wrong here ?

#39
$
\left(\dfrac0X = 0\ ,\ X \in \mathbb{R} \setminus \{0\}\right)
$
function
is ok now ? Unfortunately no:
$\color{blue}\bold{xy=0~\implies~y=0\ ,\ X \in \mathbb{R} \setminus \{0\}}$

#41
x + y = 1
y = 1- x
function
The transformations of the origianl equation is wrong:

$\color{blue}\bold{x^2+xy=1~\implies~xy=1-x^2~\implies~y=\dfrac1x-x\ ,\ x \in \mathbb{R} \setminus \{0\}}$

#42 a function <<<<< and what about the domain?
In the 2nd sentence of the initial quaestion you are asked to find the domains of the functions. You haven't calculated a domain yet!

9. For all that are NOT functions you are asked to "find a value of x top which there corresponds more than one value of y". You do not appear to have done that.

10. Originally Posted by HallsofIvy
For all that are NOT functions you are asked to "find a value of x top which there corresponds more than one value of y". You do not appear to have done that.

thankz earboth
HallsofIvy I did not understand it, can you please explain more

11. ???

12. Originally Posted by HallsofIvy
For all that are NOT functions you are asked to "find a value of x top which there corresponds more than one value of y". You do not appear to have done that.
what NOT functions ?

13. Originally Posted by r-soy
what NOT functions ?
Please check your own notes: #34 and #35 are NOT functions. So you are asked to find at least one x-value with which you get at least 2 y-values.