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Differential Equations 18 Online
OpenStudy (anonymous):

Please help!!! Simplify the rational expression. State any restrictions on the variable

OpenStudy (anonymous):

OpenStudy (gorv):

k^2-4k-5 k^2-5k+k-5 k(k-5)+(k-5)=(k-5)(k+1) it is denominator

OpenStudy (gorv):

denominator should not be equal to 0 ...bcoa it will make expression infinity as anything/0= infinity \[k \neq 5 ; k \neq-1\]

OpenStudy (gorv):

k^2-k-2 k^2-2k+k-2 =k(k-2)+(k-2) =(k-2)(k+1) numerator

OpenStudy (gorv):

\[\frac{ (k-2) (k+1)}{ (k-5)(k+1) }=\frac{ k-2 }{ k-5 }\]

OpenStudy (gorv):

as k+1 will be cancelled

OpenStudy (gorv):

\[k \neq 5\] and \[k \neq -1 \] are the restriction

OpenStudy (anonymous):

I think it's going to be either D or C just not sure which one

OpenStudy (gorv):

check the restriction ...which one macth with ours??

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