you have the function f(x)=x/5+2x-2
a) decide f(20)
f(20)=20/5+2*20-2
4+38=42
b) what is x if f(x)=31. how do i do here??
+23254 If f(x) = x/5 + 2x - 2
and f(x) = 31,
then set the two right-sides of the equations equal to each other: x/2 + 2x - 2 = 31
I would get rid of the fractions first; I would multiply each term by 2: x + 4x - 4 = 62
Then, simplify the left side: 5x - 4 = 62
Add 4 to both sides: 5x = 66
Divide by 5: x = 66/5
+23254 If f(x) = x/5 + 2x - 2
and f(x) = 31,
then set the two right-sides of the equations equal to each other: x/2 + 2x - 2 = 31
I would get rid of the fractions first; I would multiply each term by 2: x + 4x - 4 = 62
Then, simplify the left side: 5x - 4 = 62
Add 4 to both sides: 5x = 66
Divide by 5: x = 66/5
