# I want check my answer and solve some question

• Oct 10th 2009, 06:45 AM
r-soy
I want check my answer and solve some question
I want check my answer and solve some question

Q :

Attachment 13284

my answer from 33 - 38

Attachment 13285

please help me in check this answer and solve questions from 39 to 42

thanks
• Oct 10th 2009, 07:06 AM
earboth
Quote:

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

Q :

Attachment 13284

my answer from 33 - 38

Attachment 13285

please help me in check this answer and solve questions from 39 to 42

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
• Oct 10th 2009, 07:56 AM
r-soy
Quote:

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

Attachment 13288

earboth also i want solve questions from 39 to 42

thanks >>
• Oct 10th 2009, 11:09 AM
earboth
Quote:

Originally Posted by r-soy
thanks ,,,

check it now

Attachment 13288

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 (Clapping)
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.

Quote:

...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.
• Oct 10th 2009, 12:33 PM
r-soy
Now

i try :::: (Worried)
Attachment 13291
• Oct 10th 2009, 11:08 PM
earboth
Quote:

Originally Posted by r-soy
Now

i try :::: (Worried)
Attachment 13291

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 (Clapping)

#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 (Worried)
• Oct 11th 2009, 02:21 AM
r-soy
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
• Oct 11th 2009, 04:48 AM
earboth
Quote:

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

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

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

#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!
• Oct 11th 2009, 05:41 AM
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.
• Oct 11th 2009, 07:25 AM
r-soy
Quote:

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 (Crying)
• Oct 11th 2009, 11:50 AM
r-soy
???
• Oct 12th 2009, 01:08 AM
r-soy
Quote:

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 ? (Itwasntme)
• Oct 12th 2009, 04:22 AM
earboth
Quote:

Originally Posted by r-soy
what NOT functions ? (Itwasntme)

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.