A segment has one endpoint at the origin, O, and the other endpoint at A(12, 36). If point B is located on
OA such that =3 then which of the following are the coordinates of B?

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

answer: it's d

step-by-step explanation:

on plase n

the given growth  function for the length of the fish (cm) is

f(m) = 5(1.07)m = 5.35m

where 

m = months

at the end of the study, the length is 9.19 cm. the  time for conducting the study is determined as

5.35m = 9.19

m = 1.7178 months

part a.

the study was performed over a time period  of about 1.72 months. therefore  the domain for  plotting the function is 0 < = m < = 1.72.

note:

if growth  continues at this rate for up to m months, then the domain would be 0< =m< =m.

part b.

the growth function is a linear function.

when m=0, f=0.

therefore the initial growth is zero, which represents the y-intercept.

part c.

assume that the growth function is valid for 0< =m< =9.

the average rate of change of the function is the slope of the linear function, which is 5.35 cm/month.