Heather is solving an equation by graphing the expressions on either side of the equal signs. If the graph intersects at an infinite number of points, which statement could describe the equation Heather is attempting to solve? One side of the equation is a constant, and the other side of the equation is a linear expression. One side of the equation is a linear expression, and the other side of the equation is a quadratic expression. One side of the equation is a constant, and the other side of the equation is a quadratic expression. One side of the equation is a quadratic expression, and the other side of the equation is a quadratic expression.

D. one side of the equation is a quadratic expression, and the other side of the equation is a quadratic expression.

Step-by-step explanation:

by definition, an arc subtended by a central angle of some given measure has the same measure. so .

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!