Which function has a domain of x greater-than-or-equal-to 5and a range of y less-than-or-equal-to 3? y = StartRoot x minus 5 EndRoot + 3
y = StartRoot x + 5 EndRoot minus 3
y = negative StartRoot x minus 5 EndRoot + 3
y = negative StartRoot x + 5 EndRoot minus 3

The function that has a domain of 'x' greater-than-or-equal-to 5 and a range of 'y' less-than-or-equal-to 3 is;

.

Step-by-step explanation:

The domain of a function are the possible 'x' values of the function

The range of a function are the possible 'y' values of the function

The given possible functions are;

1)

The range of values for the above function are;

Domain; x ≥ 5

Range; y ≥ 3

2)

The range of values for the above function are;

Domain; x ≥ -5

Range; y ≥ -3

3)

The range of values for the above function are;

Domain; x ≥ 5

Range; y ≤ 3

4)

The range of values for the above function are;

Domain; x ≥ -5

Range; y ≤ -3

Therefore, the required function is the third function, .

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: