1. ## algebra fraction

$

pi^2/-s^3-2 = - (pi^2/-s^3-2)

$

$
= pi^2/ -s^3+2
$

am i using the subtraction correctly?

2. Hello,

I don't know what you mean by /- but I assume you left out some parentheses.

pi^2/(-s^3-2)= $\frac{pi^2}{-s^3-2}=-\frac{pi^2}{s^3+2}$.
pi^2/(-s^3)-2= $\frac{pi^2}{-s^3}-2=-\frac{pi^2}{s^3}-2$.
-(pi^2/(-s^3-2))= $-\frac{pi^2}{-s^3-2}=\frac{pi^2}{s^3+2}$.
-(pi^2/(-s^3)-2))= $-\left(\frac{pi^2}{-s^3}-2\right)=\frac{pi^2}{s^3}+2$.

You wrote pi^2/-s^3-2= pi^2/ -s^3+2 which seems obviously wrong.

Bye.

3. I'm not sure how your problem looks like. Is it

$\frac{\pi^2}{-s^3-2}$

or

$\frac{\pi^2}{-s^3} - 2$

4. ok i changed it...didnt know it would come out like that on here..

the 2nd line is the right side

im just checking if i distributed the negative right or wrong

5. Hello,

$-(A+B)=-A-B, -(A-B)=-A+B,-(-A-B)=A+B$,
$-\frac{A}{B}=\frac{A}{-B}=\frac{-A}{B}$.
Thus,
$-\frac{A}{B+C}=\frac{A}{-(B+C)}=\frac{A}{-B-C}$,
$-\frac{A}{B-C}=\frac{A}{-(B-C)}=\frac{A}{-B+C}$,
$-\frac{A}{-B-C}=\frac{A}{-(-B-C)}=\frac{A}{B+C}$.

I hope it helps. I wrote all the possibilities in the previous post.

Bye.

6. is that correct ??

7. If you saying the left hand side equals the bottom right hand side then nope.

You can see this is not true by plugging in a value for s. Try plugging in 1 and you will notice

$\frac{\pi^2}{-(1)^3 - 2}~\text{does not equal}~ \frac{\pi^2}{-(1)^3+2}$

8. so i distributed the negative correctly or no?

i just need to know if i did or not...because im checking if functions are even, odd, or neither...

the function is h(s) = [ pi^2 over s^3 - 2 ]

9. Testing to see if it even

$y = \frac{\pi^2}{-s^3-2}$

Replace s with -s

$y = \frac{\pi^2}{-(-s)^3-2}$

which equals

$y = \frac{\pi^2}{s^3-2}$

so it is not the same as the original equation so it is not even.

To see if it odd replace y with -y and s with -s

$-y = \frac{\pi^2}{-(-s)^3-2}$

$-y = \frac{\pi^2}{s^3-2}$

$y = \frac{\pi^2}{-s^3+2}$

which is not the same as the original function so it is neither

10. the original function itself h(s) does not have -s^3 - 2...

its s^3 - 2

11. Originally Posted by realintegerz

is that correct ??
Wait I thought the left hand side equation was the original equation?

So it is

$\frac{\pi^2}{s^3-2}$

12. yep..

like i said before

the function h(s) = pi^2/s^3-2

i put -s into it, and it isnt the same, so its not even...

and the picture is the work to see if its odd...

13. h(s) = pi^2/s^3 - 2
h(-s) = pi^2/(-s)^3 - 2
= pi^2/-s^3 - 2

so h(s) doesnt = h(-s)

then i tried this

trying to see if h(-s) = - h(s)
pi^2/-s^3 - 2 = - ( pi^2/ s^3 - 2)
= pi^2/-s^3 + 2

thats all my work

$y = \frac{\pi^2}{s^3-2}$

To see if it odd replace y with -y and s with -s

$-y = \frac{\pi^2}{(-s)^3-2}$

$-y = \frac{\pi^2}{-s^3-2}$

$y = \frac{\pi^2}{s^3+2}$

So nope it is not odd either