Lesson 6.2: applications of linear systems this is holding me back from completing the course. Please help me guys and girls.

see explanation

Step-by-step explanation:

When x = 5 then x > 3 , thus

f(5) = - 1

When x = 3 then x ≤ 3 , thus

f(3) = - 2x + 2 = - 2(3) + 2 = - 6 + 2 = - 4

When x = - 4 then - 4 ≤ x ≤ 1 , thus

f(- 4) = 2

When x = 3 then x > 1 , thus

f(3) = - x + 1 = - 3 + 1 = - 2

Step-by-step explanation:

-2x+2 , x≤ 3

f(x)=-1 , x>3

f(5) = -1 since 5 is greater than 3 you have to use the second equation where x>3

f(3)=-2x+3 use the second equation because the value of x is ≤3

f(3)=-6+3=-3

second question:

-x-4   when x<-4

f(x)= 2 when -4≤x≤1

f(x)=-x+1  when x>1

f(-4) use the second equation because x=-4

f(-4)=2

f(3)=-3+1=-2 use the third equation because x>1 and 3 is greater than 1

answer: the anser is 3

step-by-step explanation:

c- 2o is equal to 9