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anonymous
 3 years ago
Please help medals/fan will be given!
anonymous
 3 years ago
Please help medals/fan will be given!

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01. If f(x) = 3x + 1 and g(x) = 2x − 4, find f[g(2)]. 25 4 zero 1 2. If f(x) = −x − 1 and g(x) = 4x − 1, find f[g(−2)] −10 −9 −13 8 3. If f(x) = x + 3 and g(x) = −4x − 1, find g[f(4)]. 21 −21 19 −19 4. If f(x) = 4x + 1 and g(x) = −x − 2, find (f + g)(−1). −4<==== 4 −6 −8 5. If f(x) = 2x − 2 and g(x) = 3x − 1, find (f − g)(4). 17 −17 5 −5 6. If f(x) = x − 2 and g(x) = −2x − 1, find (f g)(9) 19 −18 zero −19 7. If f(x) = −2x − 4 and g(x) = x + 4, find ()(−8). −3 3 −1 1 8. If f(x) = x − 4 and g(x) = x − 2, find f[g(−6)]. 9 3 −9 −3 9. If h(x) = 3x − 1 and j(x) = −2x, find h[j(2)]. 11 −11 13 −13<==== 10. If r(x) = 3x − 1 and t(x) = −4x − 1, find (r + t)(7). 9 −9<==== −49 49

hba
 3 years ago
Best ResponseYou've already chosen the best response.0Please post a single question at a time.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok can you give an answer to at least one please

hba
 3 years ago
Best ResponseYou've already chosen the best response.0Giving direct Answers is not allowed on OS.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01 for the first question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0to get that you plug in 0 into g(x) then plug your solution 0 into f(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I didn't say don't explain

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you @marvin.cling

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i explained do you want more explanation? \[g(2)=2(2)4=0, f(g(2))= 3(0)+1=1. How's that for an explanation\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry i meant two not 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no just plaese answer my next post
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