The Point (0,0) is a solution to which of these inequalities?

(0, 0) s the solution of y-4 < 3x-1 as it satisfies the inequality.

Hence, option C is true.

Step-by-step explanation:

Given the point (0, 0)

a)

Putting the point (0, 0) the inequality

y+4 < 3x-1

0+4 < 3(0)-1

4 < -1

This is false as -1 can not be greater than 4

b)

y-1 < 3x-4

Putting the point (0, 0) the inequality

0-1 < 3(0)-4

-1 < -4

This is false as -1 can not be lesser than -4

c)

y-4 < 3x-1

Putting the point (0, 0) the inequality

0-4 < 3(0)-1

-4 < -1

This is true as -4 is lesser than -1

d)

y+4 < 3x+1

Putting the point (0, 0) the inequality

0+4 < 3(0)+1

4 < 1

This is false as 4 can not be lesser than 1

Therefore, (0, 0) s the solution of y-4 < 3x-1 as it satisfies the inequality.

Hence, option C is true.

answer

106 emergencies

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!