am i using the subtraction correctly?
Last edited by realintegerz; Sep 21st 2008 at 09:41 PM.
Follow Math Help Forum on Facebook and Google+
Hello, I don't know what you mean by /- but I assume you left out some parentheses. pi^2/(-s^3-2)= . pi^2/(-s^3)-2= . -(pi^2/(-s^3-2))= . -(pi^2/(-s^3)-2))= . You wrote pi^2/-s^3-2= pi^2/ -s^3+2 which seems obviously wrong. Bye.
I'm not sure how your problem looks like. Is it or
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
Hello, , . Thus, , , . I hope it helps. I wrote all the possibilities in the previous post. Bye.
is that correct ??
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
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 ]
Testing to see if it even Replace s with -s which equals 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 which is not the same as the original function so it is neither
the original function itself h(s) does not have -s^3 - 2... its s^3 - 2
Originally Posted by realintegerz is that correct ?? Wait I thought the left hand side equation was the original equation? So it is
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...
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
Ok sorry I misread To see if it odd replace y with -y and s with -s So nope it is not odd either
View Tag Cloud