Solution:
As we given that
then using this conversion factor we can write 
and further it can be written as

Hence the required conversion factor is 2.54 cm/1 inch as we can see from the above calculation.
Answer:
|F net| = 20.22 N
θ ≈ 19.8°
Step-by-step explanation:
F net = 15N i + 8cos(60°)N i + 8sin(60°)N j
= 15N i + 8×½N i + 8×√3/2N j
= 15N i + 4N i + 4√3N j
= 19N i + 4√3N j
|F net| = √(19²+(4√3)²) = √(361+48) = √409 ≈ 20.22N
tan(θ) = 4√3 ÷ 19 ≈ 0.36 → θ ≈ arctan(0.36) = 19.8°
I = / r where I = current and r = resistance
80 = k / 50 so
k = 400
so we have I = 400/r
when r = 40
I = 400/40 = 10 amps
30x - 20 = 50
use inverse operations to isolate x
30x - 20 = 50
+ 20 +20
30x = 70
(30x = 70) /30
x= 7/3 (this is the simplified term)