anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@DanJS
anonymous
  • anonymous
@DullJackel09
anonymous
  • anonymous
@Donblue

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More answers

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

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