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

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- Oct 10th 2009, 06:45 AMr-soyI 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 AMearboth
- Oct 10th 2009, 07:56 AMr-soy
thanks ,,,

check it now

Attachment 13288

earboth also i want solve questions from 39 to 42

thanks >> - Oct 10th 2009, 11:09 AMearboth
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

- Oct 10th 2009, 12:33 PMr-soy
Now

i try :::: (Worried)

Attachment 13291 - Oct 10th 2009, 11:08 PMearboth
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 AMr-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 AMearboth
- Oct 11th 2009, 05:41 AMHallsofIvy
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 AMr-soy
- Oct 11th 2009, 11:50 AMr-soy
???

- Oct 12th 2009, 01:08 AMr-soy
- Oct 12th 2009, 04:22 AMearboth