Can i get some help please this is pretty hard

1. 2/3

2. doesn't exist

3. 2

4. x ≥ 0

5. y ≥ 2

Step-by-step explanation:

If we were to take any two points on the line and divide the difference of the y-coordinates by the difference of the x-coordinates, we'd get the slope. So, take any two points: (0, 2) and (3, 4). The slope is:

(4 - 2) / (3 - 0) = 2/3

The x-intercept is where the line crosses the x-axis, but since the line only starts at (0, 2), the x-intercept doesn't exist.

The y-intercept is where the line crosses the y-axis, which is at point (0, 2) so the y-intercept is 2.

The domain is the possible x-values. Here, because the smallest x-value is 0 (from the point (0,2)), and there is no largest x-value (the arrow at the end of the line indicates the graph goes on to infinity), the domain is: x ≥ 0.

The range is the possible y-values. Here, the smallest possible y value is 2 (from (0, 2)) and the largest is infinity, so the range is: y ≥ 2.

If you do this step by step, you will see it is actually easy! So first, the slope is found by rise over run!

rise = 2; run = 3

Slope = 2/3

The graph doesn't extend to the x axis!

X intercept = Not there

The y intercept is where the function touches the y axis, this is already evident on the graph!

Y intercept = (0,2)

The domain is all possible x values, the function starts at (0,2) and can't go past that point which makes our domain...

Domain =  

The range is all possible y values, and since the function's vertical limit is 2, then our range is...

Range =  

damm bro the struggle

step-by-step explanation:

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!