Mathematics
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OpenStudy (anonymous):
please help Given the following functions f(x) and g(x), solve (f + g)(3) and select the correct answer below:
f(x) = 6x + 3
g(x) = x − 7
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OpenStudy (anonymous):
@DanJS
OpenStudy (anonymous):
@DullJackel09
OpenStudy (anonymous):
@Donblue
OpenStudy (anonymous):
@queen-of-tokyo
jimthompson5910 (jim_thompson5910):
are you able to evaluate f(x) when x = 3 ?
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OpenStudy (anonymous):
yes its 21 right?
OpenStudy (anonymous):
are u there?
jimthompson5910 (jim_thompson5910):
correct, f(3) = 21
jimthompson5910 (jim_thompson5910):
how about the value of g(3) ?
OpenStudy (anonymous):
-4?
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jimthompson5910 (jim_thompson5910):
very good
jimthompson5910 (jim_thompson5910):
now just add the values of f(3) and g(3)
jimthompson5910 (jim_thompson5910):
this works because
(f+g)(x) = f(x) + g(x)
(f+g)(3) = f(3) + g(3)
OpenStudy (anonymous):
17 times 3?
jimthompson5910 (jim_thompson5910):
add
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OpenStudy (anonymous):
4
17
25
31
jimthompson5910 (jim_thompson5910):
(f+g)(3) = f(3) + g(3)
(f+g)(3) = 21 + (-4)
OpenStudy (anonymous):
those are the choices
jimthompson5910 (jim_thompson5910):
what do you mean by 17 times 3 ?
OpenStudy (anonymous):
because 21 +-4 is 17
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jimthompson5910 (jim_thompson5910):
which is your answer
jimthompson5910 (jim_thompson5910):
I'm not sure why you are multiplying by 3
OpenStudy (anonymous):
because( f+g)(3
OpenStudy (anonymous):
u have to multiply by 3 after u add em right?
jimthompson5910 (jim_thompson5910):
the notation `(f+g)(3)` does NOT mean `(f+g) times 3`
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jimthompson5910 (jim_thompson5910):
f(3) does not mean f times 3
it means "replace every x with 3, then evaluate"
OpenStudy (anonymous):
ok could u tell me what it means
OpenStudy (anonymous):
o ok
jimthompson5910 (jim_thompson5910):
(f+g)(3) is just another function that is evaluated at x = 3
OpenStudy (anonymous):
so its 17?
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jimthompson5910 (jim_thompson5910):
yes, that is correct
OpenStudy (anonymous):
can u help with 1 more?
jimthompson5910 (jim_thompson5910):
sure
OpenStudy (anonymous):
ill create new thread then tagu
jimthompson5910 (jim_thompson5910):
ok