Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

If , f(x)= 3/(x+2) minus the square root of x-3, complete the following statement: f(19) = _____

OpenStudy (anonymous):

@driftracer305 @ramit.dour

OpenStudy (anonymous):

what you basically have to do is replace the x with 19

OpenStudy (anonymous):

so it would be f(19)=3/(19+2)-sqr root of 3.

OpenStudy (anonymous):

yes, i just want to make sure my answer is correct

OpenStudy (astrophysics):

\[f(x) = \frac{ 3 }{ (x+2) } - \sqrt{x-3}\] \[f(19) = \frac{ 3 }{ (19+2) } - \sqrt{19-3}\] evaluate this.

OpenStudy (anonymous):

I got 5 :P

OpenStudy (anonymous):

actually negative 5 so yeah -5

OpenStudy (anonymous):

hmm no i didn't get -5

OpenStudy (astrophysics):

Well just to make sure is the equation I wrote out correct? Or is the - squareroot (x-3) in the denominator.

OpenStudy (anonymous):

the equation you wrote out is correct

OpenStudy (astrophysics):

Ok well I got -27/7 then, which is -3.86.

OpenStudy (anonymous):

same! thank you

OpenStudy (astrophysics):

Np ^.^

OpenStudy (driftracer305):

wow..... i am late to the party........ sorry i was busy

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
Mari103: How to pop out like a Jacc In the box
28 minutes ago 0 Replies 0 Medals
Breathless: Spooky witch but cute
7 hours ago 3 Replies 0 Medals
Arriyanalol: help
7 hours ago 10 Replies 2 Medals
Arriyanalol: @tinydinoUwU stop trying to find a argument u blad lil boy
1 day ago 5 Replies 4 Medals
Jaded012023: Please tell me what you all think of this song
10 hours ago 6 Replies 1 Medal
Arriyanalol: bro how
10 hours ago 2 Replies 3 Medals
Arriyanalol: cant wait for the new bluey movie in 2027
1 day ago 12 Replies 2 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!