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anonymous
 one year ago
Use the functions m(x) = 5x + 4 and n(x) = 6x − 9 to complete the function operations listed below.
Part A: Find (m + n)(x). Show your work. (3 points)
Part B: Find (m ⋅ n)(x). Show your work. (3 points)
Part C: Find m[n(x)]. Show your work. (4 points)
anonymous
 one year ago
Use the functions m(x) = 5x + 4 and n(x) = 6x − 9 to complete the function operations listed below. Part A: Find (m + n)(x). Show your work. (3 points) Part B: Find (m ⋅ n)(x). Show your work. (3 points) Part C: Find m[n(x)]. Show your work. (4 points)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so basically what part A is saying is find m(x) + n(x) part B is saying m(x) * n(x) part C is saying plug n(x) equation for x in the m(x) equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for part A do i just use the number for m that does not have a x in it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so basically its saying (5x+4) + (6x9)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0\(\large { (m+n)(x)\implies m(x)+n(x) \\ \quad \\ (m\cdot n)(x)\implies m(x)\cdot n(x) \\ \quad \\ m[n(x)]\implies \begin{cases} m(x) = 5x + 4\\{\color{brown}{ n(x) }} = 6x  9 \end{cases}\qquad m[n(x)]=5[{\color{brown}{ n(x) }}]+4 }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0idk if that answers ur quesiton

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg no way so did I :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what do i put for part A?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer which u asked me about

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tell me what you get read over my first comment and try it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would i just do the same thing but multiply them instead of adding?
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