Which is more reasonable measurement of the distance between the track on a cd 1.6 x 10 to the negative third mm or 1.6 * 10^3 mm justify your answer

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10

the function makes a strait line horizontally with no x intercept but a y intercept at 9.