Brian set his compass equal to the radius of circle C and drew two circles centered at points A and B on circle C. He labeled the points of intersection of the two circles as shown. To complete his construction, Brian only needs to use his straightedge to draw some chords of circle C. Which figures could Brian be constructing? equilateral triangle MNQ inscribed in circle C equilateral triangle ANP inscribed in circle C regular hexagon AMNBPQ inscribed in circle C square MNPQ inscribed in circle C square ANBQ inscribed in circle C

AMNBPQ inscribed in circle C square MNPQ inscribed in circle C

Step-by-step explanation:

AMNBPQ inscribed in circle C square MNPQ inscribed in circle C AMNBPQ inscribed in circle C square MNPQ inscribed in circle C

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!