Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

Answer to if f(x)=-2x+8 and g(x)=√x+7, what is (f*g)(2)?

OpenStudy (anonymous):

\[\sqrt{(-2x+8)+7}\] feeling pretty good of that one

OpenStudy (anonymous):

i don't remember the (FoG) formula. can someone please remind me?

OpenStudy (anonymous):

to find (f o g) you take your x in g and replace it with the function f if i remember correctly

OpenStudy (anonymous):

why are there parenthesis?

OpenStudy (anonymous):

just to show you the substitution made

OpenStudy (lalaly):

g(2)=3 f(3)=-2(3)+8=2

OpenStudy (anonymous):

So if I remove them it's still fine?

OpenStudy (anonymous):

@lalay that looks better, what did you do?

OpenStudy (anonymous):

awww 8(

OpenStudy (anonymous):

nothing against you or your answer, I just don't think that's correct, and I don't want to submit it and find out if it is or isn't unless everyone is sure. And I haven't had a radicle at all before for these problems.. I'm sorry, idk which is right.

OpenStudy (anonymous):

I can go ahead and check though.. Might as well.

OpenStudy (lalaly):

first g(2) = sqrt 2+7 = sqrt 9 =3 then f(g(2))=f(3) = -2*3 +8 =2

OpenStudy (anonymous):

Idk what to do :(

OpenStudy (anonymous):

The first answer was incorrect.. I have a new question now?

OpenStudy (lalaly):

(f*g)(2) = f(g(2)) so first u find g(2) u substitiute 2 in g(x) which is sqrt 9 =3 then f(g(2)) = f(3) u substitute 3 in f(x) whats ur question

OpenStudy (anonymous):

If f(x)=-7x+2 and g(x)=√x+3, what is (f*g)(-2)

OpenStudy (lalaly):

g(-2) = sqrt 1 = 1 f(g(-2)) = f(1) = -7+2 = 5

OpenStudy (anonymous):

How do I solve f(g(-2)) = f(1) = -7+2 = 5?

OpenStudy (lalaly):

since g(-2) = 1 then f(g(-2)) = f(1) u jus replace g(-2) bye its value which is 1 thn substitute 1 in f(x) u get -7(1)+2 = -5

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!
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!